User:Egm4313.s12.team10.R2

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Contents
=1 Problem 1=

1.1 Given

1.2 Find

1.3 Solution

1.4 Author

=2 Problem 2=

2.1 Given

2.2 Find

2.3 Solution

2.4 Author

=3 Problem 3=

3.1 Given

3.1.1 (3)

3.1.2 (4)

3.2 Find

3.3 Solution

3.3.1 (3)

3.3.2 (4)

3.4 Author

=4 Problem 4=

4.1 Given

4.1.1 4a

4.1.2 4b

4.2 Find

4.3 Solution

4.3.1 4a

4.3.2 4b

4.4 Author

=5 Problem 5=

5.1 Given

5.1.1 5a

5.1.2 5b

5.2 Find

5.3 Solution

5.3.1 5a

5.3.2 5b

5.4 Author

=6 Problem 6: Spring-Dashpot-Mass System=

6.1 Given

6.2 Find

6.3 Solution

6.4 Author

=7 Problem 7: Develop the Maclaurin Series=

7.1 Given

7.1.1 A

7.1.2 B

7.1.3 C

7.2 Find

7.3 Solution

7.3.1 A

7.3.2 B

7.3.3 C

7.4 Author

=8 Problem 8=

8.1 Given

8.1.1 A

8.1.2 B

8.2 Find

8.3 Solution

8.3.1 A

8.3.2 B

8.4 Author

=9 Problem 9=

9.1 Given

9.2 Find

9.3 Solution

9.4 Author

=10 Contributing Members=

= Problem 1 =

Given
The following roots and initial conditions for a differential equation are given:


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 * $$\displaystyle \lambda _1 = -2, \lambda _2  = +5 $$
 * $$\displaystyle \lambda _1 = -2, \lambda _2  = +5 $$


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 * }


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 * $$\displaystyle y(0) = 1,y'(0) = 0 $$
 * $$\displaystyle y(0) = 1,y'(0) = 0 $$


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 * }

Find
For the conditions given above:

a) Solve for the standard form of the non-homogeneous, second-order ordinary differential equation with constant coefficients (L2-ODE-CC).

b) Rewrite the solution from part (a) using the initial conditions given and the general excitation r(x).

c) Plot the equation found in part (b) when r(x)=0.

d) Write 3 non-standard, non-homogeneous L2-ODE-CC that have the two lambda values above as the roots of their characteristic equations.

Solution
 Part (a) 


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 * $$\displaystyle y_{h1} (x) = e^{\lambda_1 x} = e^{-2 x}  $$
 * $$\displaystyle y_{h1} (x) = e^{\lambda_1 x} = e^{-2 x}  $$


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 * }


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 * $$\displaystyle y_{h2} (x) = e^{\lambda_2 x} = e^{5 x} $$
 * $$\displaystyle y_{h2} (x) = e^{\lambda_2 x} = e^{5 x} $$


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 * }


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 * $$\displaystyle y_h (x) = C_1 y_{h1} + C_2 y_{h2}  $$
 * $$\displaystyle y_h (x) = C_1 y_{h1} + C_2 y_{h2}  $$


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 * }


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 * $$\displaystyle y_h (x) = C_1 e^{-2x} + C_2 e^{5x} $$
 * $$\displaystyle y_h (x) = C_1 e^{-2x} + C_2 e^{5x} $$


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 * }


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 * $$\displaystyle y (x) = y_h (x) + y_p (x) $$
 * $$\displaystyle y (x) = y_h (x) + y_p (x) $$


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Standard form of the solution:


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 Part (b) 

Inputting the initial conditions given into the final equation from part (a):


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 * $$\displaystyle y(0)=1=C_1 e^{-2(0)} + C_2 e^{5(0)} + y_p (0) $$
 * $$\displaystyle y(0)=1=C_1 e^{-2(0)} + C_2 e^{5(0)} + y_p (0) $$


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 * }


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 * $$\displaystyle 1= C_1 + C_2  + y_p (0) $$
 * $$\displaystyle 1= C_1 + C_2  + y_p (0) $$


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 * }


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 * $$\displaystyle C_1 = 1- C_2 - y_p (0) $$
 * $$\displaystyle C_1 = 1- C_2 - y_p (0) $$


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 * }


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 * $$\displaystyle y'(x) = -2C_1 e^{-2x} +5C_2 e^{5x} +y'_p (x) $$
 * $$\displaystyle y'(x) = -2C_1 e^{-2x} +5C_2 e^{5x} +y'_p (x) $$


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 * }


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 * $$\displaystyle y'(0) = 0 = -2C_1 e^{-2(0)} + 5C_2 e^{5(0)} +y'_p (0) $$
 * $$\displaystyle y'(0) = 0 = -2C_1 e^{-2(0)} + 5C_2 e^{5(0)} +y'_p (0) $$


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 * }


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 * $$\displaystyle 0 = -2C_1 + 5C_2 + y'_p (0) $$
 * $$\displaystyle 0 = -2C_1 + 5C_2 + y'_p (0) $$


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 * }


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 * $$\displaystyle 5C_2 = 2C_1 - y'p(0) $$
 * $$\displaystyle 5C_2 = 2C_1 - y'p(0) $$


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 * }


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 * $$\displaystyle 5C_2 = 2[1-C_2 - y_p(0)] - y'p(0) $$
 * $$\displaystyle 5C_2 = 2[1-C_2 - y_p(0)] - y'p(0) $$


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 * }


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 * $$\displaystyle 5C_2 = 2- 2C_2 - 2y_p(0) - y'p(0) $$
 * $$\displaystyle 5C_2 = 2- 2C_2 - 2y_p(0) - y'p(0) $$


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 * }


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 * $$\displaystyle 7C_2 = 2 - 2y_p(0) - y'p(0) $$
 * $$\displaystyle 7C_2 = 2 - 2y_p(0) - y'p(0) $$


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 * }


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 * $$\displaystyle C_2 = \frac{1}{7} [2 - 2y_p(0) - y'p(0)] $$
 * $$\displaystyle C_2 = \frac{1}{7} [2 - 2y_p(0) - y'p(0)] $$


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 * }


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 * $$\displaystyle C_1 = 1 - \frac{1}{7} [2 - 2y_p(0) - y'p(0)] - y_p(0) $$
 * $$\displaystyle C_1 = 1 - \frac{1}{7} [2 - 2y_p(0) - y'p(0)] - y_p(0) $$


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 * }


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 * $$\displaystyle 7C_1 = 7 - 2 + 2y_p(0) + y'p(0) - 7y_p(0) $$
 * $$\displaystyle 7C_1 = 7 - 2 + 2y_p(0) + y'p(0) - 7y_p(0) $$


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 * }


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 * $$\displaystyle 7C_1 = 5 - 5y_p(0) + y'p(0) $$
 * $$\displaystyle 7C_1 = 5 - 5y_p(0) + y'p(0) $$


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 * }


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 * $$\displaystyle C_1 = \frac{1}{7}[5 - 5y_p(0) + y'p(0)] $$
 * $$\displaystyle C_1 = \frac{1}{7}[5 - 5y_p(0) + y'p(0)] $$


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 * }

Solution to the ODE in terms of the initial conditions given:


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 Part (c) 

We will consider the following condition:


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 * $$\displaystyle y'' + ay' + by = r(x) = 0 $$
 * $$\displaystyle y'' + ay' + by = r(x) = 0 $$


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 * }

Since r(x) is equal to zero, a particular solution would be redundant


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 * $$\displaystyle y_p(x) = 0, y'_p(x) = 0, y''_p(x) = 0 $$
 * $$\displaystyle y_p(x) = 0, y'_p(x) = 0, y''_p(x) = 0 $$


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 * }


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 * $$\displaystyle \overline{y} = y_h  + y_p $$
 * $$\displaystyle \overline{y} = y_h  + y_p $$


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 * }


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 * $$\displaystyle y(x) = \frac{1}{7} [5 + 0 - 0] e^{-2x} + \frac{1}{7}[2 - 0 - 0] e^{5x} + 0 $$
 * $$\displaystyle y(x) = \frac{1}{7} [5 + 0 - 0] e^{-2x} + \frac{1}{7}[2 - 0 - 0] e^{5x} + 0 $$


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 * }


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The equation above is the solution to the ODE under the initial conditions given and the assumption that r(x) = 0. We can now plot this solution:



 Part (d) 

An equation of the form


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 * $$\displaystyle y'' + a y' + by = r(x) $$
 * $$\displaystyle y'' + a y' + by = r(x) $$


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 * }

Has the characteristic equation of form


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 * $$\displaystyle \lambda + a \lambda + b = 0 $$
 * $$\displaystyle \lambda + a \lambda + b = 0 $$


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 * }

Which is the characteristic equation for infinitely many differential equations when multiplied by a constant factor “m”. Rewriting this equation in terms of its roots can help us form the different L2-ODE-CC that y is a solution to:


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 * $$\displaystyle m(\lambda + \lambda _1)(\lambda + \lambda _2) = 0 $$
 * $$\displaystyle m(\lambda + \lambda _1)(\lambda + \lambda _2) = 0 $$


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 * }


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 * $$\displaystyle 2(\lambda + 2)(\lambda - 5) = 2(\lambda ^2 - 3\lambda - 10) = 2\lambda ^2 - 6 \lambda - 20 $$
 * $$\displaystyle 2(\lambda + 2)(\lambda - 5) = 2(\lambda ^2 - 3\lambda - 10) = 2\lambda ^2 - 6 \lambda - 20 $$


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 * }


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 * $$\displaystyle 5(\lambda + 2)(\lambda - 5) = 5(\lambda ^2 - 3\lambda - 10) = 5\lambda ^2 - 15 \lambda - 50 $$
 * $$\displaystyle 5(\lambda + 2)(\lambda - 5) = 5(\lambda ^2 - 3\lambda - 10) = 5\lambda ^2 - 15 \lambda - 50 $$


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 * }


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 * $$\displaystyle 3(\lambda + 2)(\lambda - 5) = 3(\lambda ^2 - 3\lambda - 10) = 3\lambda ^2 - 9 \lambda - 30 $$
 * $$\displaystyle 3(\lambda + 2)(\lambda - 5) = 3(\lambda ^2 - 3\lambda - 10) = 3\lambda ^2 - 9 \lambda - 30 $$


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 * }

These characteristic equations belong to the following L2-ODE-CC:


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 * (1.1.9)
 * }


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 * (1.1.9)
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Author
Solved and typed by - David Mercado Reviewed By - Entire Group

= Problem 2 =

Given

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 * $$\displaystyle y''-10y'+25y=0 $$
 * $$\displaystyle y''-10y'+25y=0 $$


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 * $$\displaystyle y(0)=1 $$
 * $$\displaystyle y(0)=1 $$


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 * $$\displaystyle y'(0)=0 $$
 * $$\displaystyle y'(0)=0 $$


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Find
The solution for the linear, 2nd order ordinary differential equation, and plot the solution.

Solution

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 * $$\displaystyle y''-10y'+25y=0 $$
 * $$\displaystyle y''-10y'+25y=0 $$


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 * }

The Characteristic equation for this ODE is as follows:


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 * $$\displaystyle r^2-10r+25=0 $$
 * $$\displaystyle r^2-10r+25=0 $$


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To proceed, we find the roots of the characteristic equation:


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 * $$\displaystyle (r-5)(r-5)=0 $$
 * $$\displaystyle (r-5)(r-5)=0 $$


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 * $$\displaystyle r_1=5 \, \, \, \, r_2=5 $$
 * $$\displaystyle r_1=5 \, \, \, \, r_2=5 $$


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We have a double root of the characteristic equation at r=5.

Therefore, the solution to the equation is of the homogeneous form (due to the lack of a particular solution):


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 * $$\displaystyle y_h(x)=C_1e^{5x}+xC_2e^{5x} $$
 * $$\displaystyle y_h(x)=C_1e^{5x}+xC_2e^{5x} $$


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We then use the two given initial conditions to solve for the two constants:


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 * $$\displaystyle y(0)=1 \, \, \, \,y'(0)=0 $$
 * $$\displaystyle y(0)=1 \, \, \, \,y'(0)=0 $$


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 * }


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 * $$\displaystyle y(0)=1=C_1e^{(5\cdot0)}+(0)C_2e^{(5\cdot0)} $$
 * $$\displaystyle y(0)=1=C_1e^{(5\cdot0)}+(0)C_2e^{(5\cdot0)} $$


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 * $$\displaystyle y(0)=1=C_1 $$
 * $$\displaystyle y(0)=1=C_1 $$


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 * }


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 * $$\displaystyle y'(x)=5C_1e^{5x}+(5)xC_2e^{5x}+C_2e^5x $$
 * $$\displaystyle y'(x)=5C_1e^{5x}+(5)xC_2e^{5x}+C_2e^5x $$


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 * }


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 * $$\displaystyle y'(0)=5C_1e^{(5\cdot0)}+(5)(0)C_2e^{(5\cdot0)}+C_2e^{(5\cdot0)} $$
 * $$\displaystyle y'(0)=5C_1e^{(5\cdot0)}+(5)(0)C_2e^{(5\cdot0)}+C_2e^{(5\cdot0)} $$


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 * }


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 * $$\displaystyle y'(0)=0=5C_1+C_2 $$
 * $$\displaystyle y'(0)=0=5C_1+C_2 $$


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Plugging in the value of the first constant yields:


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 * $$\displaystyle 0=5+C_2 $$
 * $$\displaystyle 0=5+C_2 $$


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 * $$\displaystyle C_2=-5 $$
 * $$\displaystyle C_2=-5 $$


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Plugging the two constants into the solution equation yields:


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Shown below is a plot of the solution to the linear 2nd order Ordinary Differential Equation:



Author
Solved and typed by - Emmett Walsh Reviewed By - Entire Group

= Problem 3 =

Given
Solve problems 3 and 4 from page 59 of Advanced Engineering Mathematics

(3)

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 * $$\displaystyle y{}''+6y{}'+8.96y=0 $$
 * $$\displaystyle y{}''+6y{}'+8.96y=0 $$


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 * }

(4)

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 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$


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 * }

Find
Find the general solution of the equation and check answer by substitution

(3)
Given Equation
 * {| style="width:100%" border="0"


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 * $$\displaystyle y{}''+6y{}'+8.96y=0 $$
 * $$\displaystyle y{}''+6y{}'+8.96y=0 $$


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 * }


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 * $$\displaystyle \frac{d^2y}{dx^2}+6\frac{dy}{dx}+8.96y=0 $$
 * $$\displaystyle \frac{d^2y}{dx^2}+6\frac{dy}{dx}+8.96y=0 $$

Solving the equation below, one can determine the type of roots are present when the answer is related to zero.
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 * $$\displaystyle a^{2}-4b $$
 * $$\displaystyle a^{2}-4b $$


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a and b vaules
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 * $$\displaystyle a=6,b=8.96 $$
 * $$\displaystyle a=6,b=8.96 $$


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Two Real Roots
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 * $$\displaystyle 6^{2}-4(8.96)>0 $$
 * $$\displaystyle 6^{2}-4(8.96)>0 $$


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 * }

General Equation
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 * $$\displaystyle y=C_{1}e^{\lambda _{1}x}+C_{2}e^{\lambda _{2}x} $$
 * $$\displaystyle y=C_{1}e^{\lambda _{1}x}+C_{2}e^{\lambda _{2}x} $$


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 * }

Plugging Lambda in for dy/dx
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 * $$\displaystyle \lambda ^{2}+6\lambda+8.96=0 $$
 * $$\displaystyle \lambda ^{2}+6\lambda+8.96=0 $$


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Solving For Lambda
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 * $$\displaystyle (\lambda +2.8)(\lambda +3.2)=0 $$
 * $$\displaystyle (\lambda +2.8)(\lambda +3.2)=0 $$


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Lambda Values
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 * $$\displaystyle \lambda =-2.8,-3.2 $$
 * $$\displaystyle \lambda =-2.8,-3.2 $$


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Final Equation
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 * (1.1.9)
 * }

Checking By Substitution

Equation
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 * $$\displaystyle y=C_{1}e^{-2.8x}+C_{2}e^{-3.2x} $$
 * $$\displaystyle y=C_{1}e^{-2.8x}+C_{2}e^{-3.2x} $$


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 * }

First Derivative
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 * $$\displaystyle y{}'=-2.8C_{1}e^{-2.8x}-3.2C_{_{2}}e^{-3.2x} $$
 * $$\displaystyle y{}'=-2.8C_{1}e^{-2.8x}-3.2C_{_{2}}e^{-3.2x} $$


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 * }

Second Derivative


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 * $$\displaystyle y{}''=7.84C_{1}e^{-2.8x}+10.24C_{_{2}}e^{-3.2x} $$
 * $$\displaystyle y{}''=7.84C_{1}e^{-2.8x}+10.24C_{_{2}}e^{-3.2x} $$


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 * }

General Equation
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 * $$\displaystyle y{}''+6y{}'+8.96y=0 $$
 * $$\displaystyle y{}''+6y{}'+8.96y=0 $$


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 * }

Plunging in Derivatives


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 * $$\displaystyle [2.8C_{1}e^{-2.8x}-3.2C_{_{2}}e^{-3.2x}]+6[7.84C_{1}e^{-2.8x}+10.24C_{_{2}}e^{-3.2x}]+8.96[C_{1}e^{-2.8x}+C_{2}e^{-3.2x}]=0 $$
 * $$\displaystyle [2.8C_{1}e^{-2.8x}-3.2C_{_{2}}e^{-3.2x}]+6[7.84C_{1}e^{-2.8x}+10.24C_{_{2}}e^{-3.2x}]+8.96[C_{1}e^{-2.8x}+C_{2}e^{-3.2x}]=0 $$


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 * }

Factoring out C values


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 * $$\displaystyle C_{1}e^{-2.8x}[7.84-16.8+8.96]+C_{2}e^{-2.8x}[10.24-19.2+8.96]=0 $$
 * $$\displaystyle C_{1}e^{-2.8x}[7.84-16.8+8.96]+C_{2}e^{-2.8x}[10.24-19.2+8.96]=0 $$


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 * }

Final result ends in zero, solution checks out.


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 * (1.1.9)
 * }

Is the correct answer

(4)
Given Equation
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 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$

Constant Values
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 * $$\displaystyle a = 4 $$
 * $$\displaystyle a = 4 $$


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 * $$\displaystyle b = (\pi^2 + 4) $$
 * $$\displaystyle b = (\pi^2 + 4) $$


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 * }


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 * $$\displaystyle \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y = 0 $$
 * $$\displaystyle \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y = 0 $$


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 * }


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 * $$\displaystyle \lambda = \frac{dy}{dx} $$
 * $$\displaystyle \lambda = \frac{dy}{dx} $$

Substituting Lambda for y Values
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 * }
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 * $$\displaystyle \lambda^2 + 4\lambda + (\pi^2 + 4) = 0 $$
 * $$\displaystyle \lambda^2 + 4\lambda + (\pi^2 + 4) = 0 $$

Solving For Omega
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 * }


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 * $$\displaystyle \lambda = - \frac{1}{2}a\, \pm i\omega $$
 * $$\displaystyle \lambda = - \frac{1}{2}a\, \pm i\omega $$


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 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \omega^2 = b-\frac{1}{4}a^2 $$
 * $$\displaystyle \omega^2 = b-\frac{1}{4}a^2 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \omega^2 = (\pi^2 + 4) - 4 $$
 * $$\displaystyle \omega^2 = (\pi^2 + 4) - 4 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \omega^2 = \pi^2 $$
 * $$\displaystyle \omega^2 = \pi^2 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \omega = \pi $$
 * $$\displaystyle \omega = \pi $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda = -2\, \pm \pi i $$
 * $$\displaystyle \lambda = -2\, \pm \pi i $$


 * style= |
 * }

Solving For y1 and y2
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_{1} = e^{\frac{-ax}{2}}\cos\omega x $$
 * $$\displaystyle y_{1} = e^{\frac{-ax}{2}}\cos\omega x $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_{2} = e^{\frac{-ax}{2}}\sin\omega x $$
 * $$\displaystyle y_{2} = e^{\frac{-ax}{2}}\sin\omega x $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_{1} = e^{-2x}\cos\pi x $$
 * $$\displaystyle y_{1} = e^{-2x}\cos\pi x $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_{2} = e^{-2x}\sin\pi x $$
 * $$\displaystyle y_{2} = e^{-2x}\sin\pi x $$


 * style= |
 * }

The general equation is:


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * (1.1.9)
 * }

Solving for our equation, we get:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x) $$
 * $$\displaystyle y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x) $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y' = {-2e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)+ e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) $$
 * $$\displaystyle y' = {-2e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)+ e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 2e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) - 2e^{-2x}(-\pi{C_1}\sin\pi{x} + \pi{C_2}\cos\pi{x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 2e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) - 2e^{-2x}(-\pi{C_1}\sin\pi{x} + \pi{C_2}\cos\pi{x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$

Pluging y", y' and y into Initial Equation
 * style= |
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) - 8e^{-2x}({C_1}\cos\pi{x}+{C_2}\sin\pi{x}) + 4e^{-2x}(-\pi{C_1}\sin\pi{x}+\pi{C_2}\cos\pi{x}) + (\pi^2 +4)[e^{-2x}({C_1}\cos\pi{x} + {C_2}\sin\pi{x})] = 0 $$
 * $$\displaystyle {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) - 8e^{-2x}({C_1}\cos\pi{x}+{C_2}\sin\pi{x}) + 4e^{-2x}(-\pi{C_1}\sin\pi{x}+\pi{C_2}\cos\pi{x}) + (\pi^2 +4)[e^{-2x}({C_1}\cos\pi{x} + {C_2}\sin\pi{x})] = 0 $$


 * style= |
 * }

Therefore:


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * (1.1.9)
 * }

is the solution

Author
Solved and typed by - Ryan Tillman,Nate Wanzie,Seth O'Brien Reviewed By - Entire Group

= Problem 4 =

4a

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$


 * style= |
 * }

4b

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle 10y''-32y'+25.6y=0 $$
 * $$\displaystyle 10y''-32y'+25.6y=0 $$


 * style= |
 * }

Find
The solution to the linear second order ordinary differential equation with constant coefficients.

Check the solution by substitution.

4a

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$


 * style= |
 * }

The ODE above can be simplified by using the variables below.


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle a = 2\pi $$
 * $$\displaystyle a = 2\pi $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle b = \pi ^2 $$
 * $$\displaystyle b = \pi ^2 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' + a y' + by = 0 $$
 * $$\displaystyle y'' + a y' + by = 0 $$


 * style= |
 * }

The equation is now in the general format for linear ODE’s.


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y = e^{\lambda x} $$
 * $$\displaystyle y = e^{\lambda x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y' = \lambda e^{\lambda x} $$
 * $$\displaystyle y' = \lambda e^{\lambda x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' = \lambda ^2 e^{\lambda x} $$
 * $$\displaystyle y'' = \lambda ^2 e^{\lambda x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda ^2 e^{\lambda x} + 2 \pi \lambda e ^{\lambda x} + \pi ^2 e ^{ \lambda x} = 0 $$
 * $$\displaystyle \lambda ^2 e^{\lambda x} + 2 \pi \lambda e ^{\lambda x} + \pi ^2 e ^{ \lambda x} = 0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle e^{\lambda x} (\lambda ^2 + 2\pi\lambda + \pi ^2) = 0 $$
 * $$\displaystyle e^{\lambda x} (\lambda ^2 + 2\pi\lambda + \pi ^2) = 0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda ^2 + 2\pi\lambda + \pi ^2 = 0 $$
 * $$\displaystyle \lambda ^2 + 2\pi\lambda + \pi ^2 = 0 $$

Solving For Lambda Values
 * style= |
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a \pi \sqrt{a^2 - 4 b}}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a \pi \sqrt{a^2 - 4 b}}{2} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{(2\pi)^2 - 4 \pi ^2}}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{(2\pi)^2 - 4 \pi ^2}}{2} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{4\pi^2 - 4 \pi ^2}}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{4\pi^2 - 4 \pi ^2}}{2} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi}{2} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a}{2} $$


 * style= |
 * }

This means that the characteristic equation has a doubleroot.


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_1 = e ^{(ax/2)} = e^{\pi x} $$
 * $$\displaystyle y_1 = e ^{(ax/2)} = e^{\pi x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_2 = u y_1 $$
 * $$\displaystyle y_2 = u y_1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle u = C_2 x + C_1, C_1 = 0, C_2 = 1 $$
 * $$\displaystyle u = C_2 x + C_1, C_1 = 0, C_2 = 1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_2 = x y_1 = x e^{-\pi x} $$
 * $$\displaystyle y_2 = x y_1 = x e^{-\pi x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * style="width:100%" |


 * style= |
 * }

This is the general solution to the ordinary differential equation.


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y' = -\pi C_1 e^{-\pi x}+C_2e^{-\pi x}-\pi C_2 x e^{-\pi x} $$
 * $$\displaystyle y' = -\pi C_1 e^{-\pi x}+C_2e^{-\pi x}-\pi C_2 x e^{-\pi x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' = \pi ^2 C_1 e^{-\pi x}-\pi C_2e^{-\pi x}+\pi ^2 C_2 x e^{-\pi x} - \pi C_2e^{-\pi x} $$
 * $$\displaystyle y'' = \pi ^2 C_1 e^{-\pi x}-\pi C_2e^{-\pi x}+\pi ^2 C_2 x e^{-\pi x} - \pi C_2e^{-\pi x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle e^{-\pi x} [\pi ^2 C_1 -\pi C_2 + \pi ^2 C_2 x - \pi - 2\pi^2 C_1 + 2\pi C_2 - 2\pi ^2 C_2 x + C_1 \pi ^2 + \pi ^2 C_2x] = 0 $$
 * $$\displaystyle e^{-\pi x} [\pi ^2 C_1 -\pi C_2 + \pi ^2 C_2 x - \pi - 2\pi^2 C_1 + 2\pi C_2 - 2\pi ^2 C_2 x + C_1 \pi ^2 + \pi ^2 C_2x] = 0 $$


 * style= |
 * }

All the terms inside the bracket cancel out and equal zero, so the general solution holds.

4b

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle 10y''-32y'+25.6y=0 $$
 * $$\displaystyle 10y''-32y'+25.6y=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \frac{d}{dx}=\lambda $$
 * $$\displaystyle \frac{d}{dx}=\lambda $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle10 \lambda^2-32\lambda+25.6=0 $$
 * $$\displaystyle10 \lambda^2-32\lambda+25.6=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle 10(\lambda-1.6)^2=0 $$
 * $$\displaystyle 10(\lambda-1.6)^2=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda-1.6=0 $$
 * $$\displaystyle \lambda-1.6=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda=1.6 $$
 * $$\displaystyle \lambda=1.6 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle R_1=1.6 $$
 * $$\displaystyle R_1=1.6 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle R_2=1.6 $$
 * $$\displaystyle R_2=1.6 $$


 * style= |
 * }

The characteristic equation has a double root at R=1.6, therefore the solution is of the form:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * style="width:100%" |


 * style= |
 * }

By using the derived equation and its derivatives we use substitution to check for confirmation of our solution:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'=1.6C_1e^{1.6x}+C_2e^{1.6x}+1.6C_2xe^{1.6x} $$
 * $$\displaystyle y'=1.6C_1e^{1.6x}+C_2e^{1.6x}+1.6C_2xe^{1.6x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y''=2.56C_1e^{1.6x}+1.6C_2e^{1.6x}+1.6C_2e^{1.6x}+2.56C_2xe^{1.6x} $$
 * $$\displaystyle y''=2.56C_1e^{1.6x}+1.6C_2e^{1.6x}+1.6C_2e^{1.6x}+2.56C_2xe^{1.6x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle 10y''-32y'+25.6y=0 $$
 * $$\displaystyle 10y''-32y'+25.6y=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle 10y''=25.6C_1e^{1.6x}+16C_2e^{1.6x}+16C_2e^{1.6x}+25.6C_2xe^1.6x $$
 * $$\displaystyle 10y''=25.6C_1e^{1.6x}+16C_2e^{1.6x}+16C_2e^{1.6x}+25.6C_2xe^1.6x $$

|| $$\displaystyle (Eq. a) $$
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle 32y'=51.2C_1e^{1.6x}+32C_2e^{1.6x}+51.2xe^{1.6x} $$
 * $$\displaystyle 32y'=51.2C_1e^{1.6x}+32C_2e^{1.6x}+51.2xe^{1.6x} $$

|| $$\displaystyle (Eq. b) $$
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle 25.6y=25.6C_1e^{1.6x}+25.6xe^{1.6x} $$
 * $$\displaystyle 25.6y=25.6C_1e^{1.6x}+25.6xe^{1.6x} $$

|| $$\displaystyle (Eq. c) $$
 * }

Combining equations a, b, and c in a manner according with that of our solution yields:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle 10y''-32y'+25.6y=0 $$
 * $$\displaystyle 10y''-32y'+25.6y=0 $$


 * style= |
 * }

Therefore the equation is confirmed via substitution.

Author
Solved and typed by - Emmett Walsh,David Mercado,Andrew Everidge Reviewed By - Entire Group

= Problem 5 =

5a
Equation form:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y''+ay'+by=0  $$
 * $$\displaystyle y''+ay'+by=0  $$


 * style= |
 * }

Roots of the characteristic equation:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle r_1=2.6 \, \, \, r_2=-4.3 $$
 * $$\displaystyle r_1=2.6 \, \, \, r_2=-4.3 $$


 * style= |
 * }

5b
Equation form:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y''+ay'+by=0 $$
 * $$\displaystyle y''+ay'+by=0 $$


 * style= |
 * }

Roots of the characteristic equation:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle r_1=-\sqrt{5} \, \, \, r_2=-\sqrt{5} $$
 * $$\displaystyle r_1=-\sqrt{5} \, \, \, r_2=-\sqrt{5} $$


 * style= |
 * }

Find
A second order linear ordinary differential equation that is of the given form.

5a

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda=\frac{d}{dx} $$
 * $$\displaystyle \lambda=\frac{d}{dx} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle (\lambda-2.6)(\lambda+4.3) $$
 * $$\displaystyle (\lambda-2.6)(\lambda+4.3) $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda^2-2.6\lambda+4.3\lambda-11.18=0 $$
 * $$\displaystyle \lambda^2-2.6\lambda+4.3\lambda-11.18=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda^2+1.7\lambda-11.18=0 $$
 * $$\displaystyle \lambda^2+1.7\lambda-11.18=0 $$


 * style= |
 * }

Working backwards from the derived characteristic equation to obtain the necessary ordinary differential equation yields:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * style="width:100%" |


 * style= |
 * }

5b

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda=\frac{d}{dx} $$
 * $$\displaystyle \lambda=\frac{d}{dx} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle (\lambda+\sqrt{5})^2  $$
 * $$\displaystyle (\lambda+\sqrt{5})^2  $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda^2+2\sqrt{5}\lambda+5=0 $$
 * $$\displaystyle \lambda^2+2\sqrt{5}\lambda+5=0 $$


 * style= |
 * }

Using the known correlation between the characteristic equation and linear second order ordinary differential equations yields our solution:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * style="width:100%" |


 * style= |
 * }

Author
Solved and typed by - Ryan Tillman,Emmett Walsh, Reviewed By - Entire Group

= Problem 6: Spring-Dashpot-Mass System =

Given


Given a spring-dashpot-mass system in series as shown in the figure above taken from lecture notes p.1-4, with real double root $$\displaystyle \lambda = -3 $$.

Find
Find the values for the parameters $$\displaystyle k, c, m $$.

Solution
The equation of motion for the above system is as follows:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle m(\overline{{y}_k''}+\frac{k}{c}\overline{y}_k)+k\overline{y}_k=\overline{f}(t) $$
 * $$\displaystyle m(\overline{{y}_k''}+\frac{k}{c}\overline{y}_k)+k\overline{y}_k=\overline{f}(t) $$

|| $$\displaystyle $$
 * }

Using the characteristic equation to find the value of the coefficients:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle (\lambda + 3)^2 = \lambda^2 + 6\lambda + 9 = 0 $$
 * $$\displaystyle (\lambda + 3)^2 = \lambda^2 + 6\lambda + 9 = 0 $$

|| $$\displaystyle $$
 * }

This yields the following ordinary differential equation:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' + 6y' + 9y = f(t) $$
 * $$\displaystyle y'' + 6y' + 9y = f(t) $$

|| $$\displaystyle $$
 * }

Solving for the coefficients:


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * (1.1.9)
 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * (1.1.9)
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle m\frac{k}{c} = 6 $$
 * $$\displaystyle m\frac{k}{c} = 6 $$

|| $$\displaystyle $$
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle 1\frac{9}{c} = 6 $$
 * $$\displaystyle 1\frac{9}{c} = 6 $$

|| $$\displaystyle $$
 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

Author
Solved and typed by - Andrew Everidge,Nate Wanzie Reviewed By - Entire Group

= Problem 7: Develop the Maclaurin Series =

A

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle e^t $$
 * $$\displaystyle e^t $$


 * style= |
 * }

B

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \sin(t) $$
 * $$\displaystyle \sin(t) $$


 * style= |
 * }

C

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \cos(t) $$
 * $$\displaystyle \cos(t) $$


 * style= |
 * }

Find
Develop the Maclaurin series for the given.

Solution
Taylor Series
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f(a) + \frac{f'(a)}{1!}(x-a)+\frac{f(a)}{2!}(x-a)^2+\frac{f'(a)}{3!}(x-a)^3+...
 * $$\displaystyle f(a) + \frac{f'(a)}{1!}(x-a)+\frac{f(a)}{2!}(x-a)^2+\frac{f'(a)}{3!}(x-a)^3+...

$$
 * style= |
 * }

Maclaurin Series when a=0
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \sum_{n=0}^{\infty}\frac{f^n(a)}{1!}(x-a)^n
 * $$\displaystyle \sum_{n=0}^{\infty}\frac{f^n(a)}{1!}(x-a)^n

$$
 * style= |
 * }

A

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f(t)=e^t \;\;\;\;\; \;\;\; f(0)=1 $$
 * $$\displaystyle f(t)=e^t \;\;\;\;\; \;\;\; f(0)=1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f'(t)=e^t \;\;\;\;\;\;\; f'(0)=1 $$
 * $$\displaystyle f'(t)=e^t \;\;\;\;\;\;\; f'(0)=1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f(t)=e^t \;\;\;\;\; f(0)=1 $$
 * $$\displaystyle f(t)=e^t \;\;\;\;\; f(0)=1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f(t)=e^t \;\;\;\; f(0)=1 $$
 * $$\displaystyle f(t)=e^t \;\;\;\; f(0)=1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

B

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f(t)=sin(t) \;\;\;\;\;\;\;\;\;\; f(0)=0 $$
 * $$\displaystyle f(t)=sin(t) \;\;\;\;\;\;\;\;\;\; f(0)=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f'(t)=cos(t) \;\;\;\;\;\;\;\; f'(0)=1 $$
 * $$\displaystyle f'(t)=cos(t) \;\;\;\;\;\;\;\; f'(0)=1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f(t)=-sin(t) \;\;\;\;\;\; f(0)=0 $$
 * $$\displaystyle f(t)=-sin(t) \;\;\;\;\;\; f(0)=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f(t)=-cos(t) \;\;\;\;\; f(0)=-1 $$
 * $$\displaystyle f(t)=-cos(t) \;\;\;\;\; f(0)=-1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f^{iv}(t)=sin(t) \;\;\;\;\;\;\;\;\; f^{iv}(0)=0 $$
 * $$\displaystyle f^{iv}(t)=sin(t) \;\;\;\;\;\;\;\;\; f^{iv}(0)=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f^{v}(t)=cos(t) \;\;\;\;\;\;\; f^{v}(0)=1 $$
 * $$\displaystyle f^{v}(t)=cos(t) \;\;\;\;\;\;\; f^{v}(0)=1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f^{vi}(t)=-sin(t) \;\;\;\;\; f^{vi}(0)=0 $$
 * $$\displaystyle f^{vi}(t)=-sin(t) \;\;\;\;\; f^{vi}(0)=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f^{vii}(t)=-cos(t) \;\;\;\; f^{vii}(0)=-1 $$
 * $$\displaystyle f^{vii}(t)=-cos(t) \;\;\;\; f^{vii}(0)=-1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \sin(t) = \frac{0}{0!} + \frac{1}{1!}(t) + \frac{0}{2!}(t^2) + \frac{-1}{3!}(t^3) + \frac{0}{4!}(t^4) + \frac{1}{5!}(t^5) + \frac{0}{6!}(t^6) + \frac{-1}{7!}(t^7) + ...
 * $$\displaystyle \sin(t) = \frac{0}{0!} + \frac{1}{1!}(t) + \frac{0}{2!}(t^2) + \frac{-1}{3!}(t^3) + \frac{0}{4!}(t^4) + \frac{1}{5!}(t^5) + \frac{0}{6!}(t^6) + \frac{-1}{7!}(t^7) + ...

$$
 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

C

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f(t)=cos(t) \;\;\;\;\;\;\;\;\;\; f(0)=1 $$
 * $$\displaystyle f(t)=cos(t) \;\;\;\;\;\;\;\;\;\; f(0)=1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f'(t)=-sin(t) \;\;\;\;\;\;\;\; f'(0)=0 $$
 * $$\displaystyle f'(t)=-sin(t) \;\;\;\;\;\;\;\; f'(0)=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f(t)=-cos(t) \;\;\;\;\;\; f(0)=-1 $$
 * $$\displaystyle f(t)=-cos(t) \;\;\;\;\;\; f(0)=-1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f(t)=sin(t) \;\;\;\;\; f(0)=0 $$
 * $$\displaystyle f(t)=sin(t) \;\;\;\;\; f(0)=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f^{iv}(t)=cos(t) \;\;\;\;\;\;\;\;\; f^{iv}(0)=1 $$
 * $$\displaystyle f^{iv}(t)=cos(t) \;\;\;\;\;\;\;\;\; f^{iv}(0)=1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f^{v}(t)=-sin(t) \;\;\;\;\;\;\; f^{v}(0)=0 $$
 * $$\displaystyle f^{v}(t)=-sin(t) \;\;\;\;\;\;\; f^{v}(0)=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f^{vi}(t)=-cos(t) \;\;\;\;\; f^{vi}(0)=-1 $$
 * $$\displaystyle f^{vi}(t)=-cos(t) \;\;\;\;\; f^{vi}(0)=-1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle f^{vii}(t)=sin(t) \;\;\;\; f^{vii}(0)=0 $$
 * $$\displaystyle f^{vii}(t)=sin(t) \;\;\;\; f^{vii}(0)=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \cos(t) = \frac{1}{0!} + \frac{0}{1!}(t) + \frac{-1}{2!}(t^2) + \frac{0}{3!}(t^3) + \frac{1}{4!}(t^4) + \frac{0}{5!}(t^5) + \frac{-1}{6!}(t^6) + \frac{0}{7!}(t^7) + ...
 * $$\displaystyle \cos(t) = \frac{1}{0!} + \frac{0}{1!}(t) + \frac{-1}{2!}(t^2) + \frac{0}{3!}(t^3) + \frac{1}{4!}(t^4) + \frac{0}{5!}(t^5) + \frac{-1}{6!}(t^6) + \frac{0}{7!}(t^7) + ...

$$
 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

Author
Solved and typed by - Silvio Attanasio,Nate Wanzie Reviewed By - Entire Group

= Problem 8 =

A

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' + y' + 3.25y = 0 $$
 * $$\displaystyle y'' + y' + 3.25y = 0 $$


 * style= |
 * }

B

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$
 * $$\displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$


 * style= |
 * }

Find
The general solution to the given differential equations

A

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' + y' + 3.25y = 0 $$
 * $$\displaystyle y'' + y' + 3.25y = 0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda^2 + \lambda + 3.25 = 0 $$
 * $$\displaystyle \lambda^2 + \lambda + 3.25 = 0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle x = \frac{-1 \pm \sqrt{1 - 4(1)(3.25)}}{(2)(1)} $$
 * $$\displaystyle x = \frac{-1 \pm \sqrt{1 - 4(1)(3.25)}}{(2)(1)} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle x = \frac{-1 \pm \sqrt{-12}}{2} $$
 * $$\displaystyle x = \frac{-1 \pm \sqrt{-12}}{2} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle x = \frac{-1 \pm i\sqrt{12}}{2} $$
 * $$\displaystyle x = \frac{-1 \pm i\sqrt{12}}{2} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_1 = \frac{-1 + i\sqrt{12}}{2} $$
 * $$\displaystyle \lambda_1 = \frac{-1 + i\sqrt{12}}{2} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_2 = \frac{-1 - i\sqrt{12}}{2} $$
 * $$\displaystyle \lambda_2 = \frac{-1 - i\sqrt{12}}{2} $$


 * style= |
 * }

Therefore the general solution to this general equation is:


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

Now, differentiating this solution to get y', and y'&apos; , we get:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y' = -\frac{1}{2}e^{\frac{-1}{2}x} (C_1\cos\sqrt{3}x + C_2\sin\sqrt{3}x) + e^{\frac{-1}{2}x}(-\sqrt{3}C_1\sin\sqrt{3}\,x + \sqrt{3}C_2\cos\sqrt{3}\, x) $$
 * $$\displaystyle y' = -\frac{1}{2}e^{\frac{-1}{2}x} (C_1\cos\sqrt{3}x + C_2\sin\sqrt{3}x) + e^{\frac{-1}{2}x}(-\sqrt{3}C_1\sin\sqrt{3}\,x + \sqrt{3}C_2\cos\sqrt{3}\, x) $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' = \frac{1}{4}e^{\frac{-1}{2}x} (C_1\cos\sqrt{3}x + C_2\sin\sqrt{3}x) - \frac{1}{2}e^{\frac{-1}{2}x}(-\sqrt{3}C_1\sin\sqrt{3}\,x + \sqrt{3}C_2\cos\sqrt{3}\, x) - \frac{1}{2}e^{\frac{-1}{2}x}(-\sqrt{3}C_1\sin\sqrt{3}\,x + \sqrt{3}C_2\cos\sqrt{3}\, x) + e^{\frac{-1}{2}x}(-12C_1\cos\sqrt{3}\,x - 12C_2\sin\sqrt{3}\, x) $$
 * $$\displaystyle y'' = \frac{1}{4}e^{\frac{-1}{2}x} (C_1\cos\sqrt{3}x + C_2\sin\sqrt{3}x) - \frac{1}{2}e^{\frac{-1}{2}x}(-\sqrt{3}C_1\sin\sqrt{3}\,x + \sqrt{3}C_2\cos\sqrt{3}\, x) - \frac{1}{2}e^{\frac{-1}{2}x}(-\sqrt{3}C_1\sin\sqrt{3}\,x + \sqrt{3}C_2\cos\sqrt{3}\, x) + e^{\frac{-1}{2}x}(-12C_1\cos\sqrt{3}\,x - 12C_2\sin\sqrt{3}\, x) $$


 * style= |
 * }

Plug in values of y, y' , and y'&apos; into the original equation.

With these values plugged in:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' + y' + 3.25y = 0 $$
 * $$\displaystyle y'' + y' + 3.25y = 0 $$


 * style= |
 * }

Still holds true.

Therefore:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y = e^{\frac{-1}{2}x} (C_1\cos\sqrt{3}x + C_2\sin\sqrt{3}x) $$
 * $$\displaystyle y = e^{\frac{-1}{2}x} (C_1\cos\sqrt{3}x + C_2\sin\sqrt{3}x) $$


 * style= |
 * }

Is the general solution to the differential equation.

B

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$
 * $$\displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda^2 + 0.54\lambda + (0.0729 + \pi) = 0 $$
 * $$\displaystyle \lambda^2 + 0.54\lambda + (0.0729 + \pi) = 0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle x = \frac{-0.54 \pm \sqrt{0.54 - 4(1)(0.0729 + \pi)}}{(2)(1)} $$
 * $$\displaystyle x = \frac{-0.54 \pm \sqrt{0.54 - 4(1)(0.0729 + \pi)}}{(2)(1)} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle x = \frac{-0.54 \pm \sqrt{-4\pi}}{2} $$
 * $$\displaystyle x = \frac{-0.54 \pm \sqrt{-4\pi}}{2} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle x = \frac{-0.54 \pm 2i\sqrt{\pi}}{2} $$
 * $$\displaystyle x = \frac{-0.54 \pm 2i\sqrt{\pi}}{2} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle x = -0.27 \pm i\sqrt{\pi} $$
 * $$\displaystyle x = -0.27 \pm i\sqrt{\pi} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_1 = -0.27 + i\sqrt{\pi} $$
 * $$\displaystyle \lambda_1 = -0.27 + i\sqrt{\pi} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_2 = -0.27 - i\sqrt{\pi} $$
 * $$\displaystyle \lambda_2 = -0.27 - i\sqrt{\pi} $$


 * style= |
 * }

Therefore the general solution to this general equation is:


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

Now, differentiating this solution to get y', and y'&apos; , we get:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y' = -0.27e^{-0.27x} (C_1\cos\sqrt{\pi}x + C_2\sin\sqrt{\pi}x) + e^{-0.27x}(-\sqrt{\pi}C_1\sin\sqrt{\pi}\,x + \sqrt{\pi}C_2\cos\sqrt{\pi}\, x) $$
 * $$\displaystyle y' = -0.27e^{-0.27x} (C_1\cos\sqrt{\pi}x + C_2\sin\sqrt{\pi}x) + e^{-0.27x}(-\sqrt{\pi}C_1\sin\sqrt{\pi}\,x + \sqrt{\pi}C_2\cos\sqrt{\pi}\, x) $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' = 0.0729e^{-0.27x} (C_1\cos\sqrt{\pi}x + C_2\sin\sqrt{\pi}x) - 0.27e^{-0.27x}(-\sqrt{\pi}C_1\sin\sqrt{\pi}\,x + \sqrt{\pi}C_2\cos\sqrt{\pi}\, x) - 0.27e^{-0.27x}(-\sqrt{\pi}C_1\sin\sqrt{\pi}\,x + \sqrt{\pi}C_2\cos\sqrt{\pi}\, x) + e^{-0.27x}(-\pi C_1\cos\sqrt{\pi}\,x - \pi C_2\sin\sqrt{\pi}\, x) $$
 * $$\displaystyle y'' = 0.0729e^{-0.27x} (C_1\cos\sqrt{\pi}x + C_2\sin\sqrt{\pi}x) - 0.27e^{-0.27x}(-\sqrt{\pi}C_1\sin\sqrt{\pi}\,x + \sqrt{\pi}C_2\cos\sqrt{\pi}\, x) - 0.27e^{-0.27x}(-\sqrt{\pi}C_1\sin\sqrt{\pi}\,x + \sqrt{\pi}C_2\cos\sqrt{\pi}\, x) + e^{-0.27x}(-\pi C_1\cos\sqrt{\pi}\,x - \pi C_2\sin\sqrt{\pi}\, x) $$


 * style= |
 * }

Plug in values of y, y' , and y'&apos; into the original equation.

With these values plugged in:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$
 * $$\displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$


 * style= |
 * }

Still holds true.

Therefore:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y = e^{-0.27x} (C_1\cos\sqrt{\pi}x + C_2\sin\sqrt{\pi}x) $$
 * $$\displaystyle y = e^{-0.27x} (C_1\cos\sqrt{\pi}x + C_2\sin\sqrt{\pi}x) $$


 * style= |
 * }

Is the general solution to the differential equation.

Author
Solved and typed by - Seth O'Brien Reviewed By - Entire Group

= Problem 9 =

Given
The ODE:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y''+4y'+13y=0 $$
 * $$\displaystyle y''+4y'+13y=0 $$


 * style= |
 * }

Initial conditions:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y(0)=1 \text{, } y'(0)=0 $$
 * $$\displaystyle y(0)=1 \text{, } y'(0)=0 $$


 * style= |
 * }

No Excitation:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle r(x)=0 $$
 * $$\displaystyle r(x)=0 $$


 * style= |
 * }

Find
Find and plot the solution to the second-order linear ODE:

Solution

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y''+4y'+13y=0 $$
 * $$\displaystyle y''+4y'+13y=0 $$


 * style= |
 * }

Corresponding characteristic equation:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda ^2+4\lambda +13=0 $$
 * $$\displaystyle \lambda ^2+4\lambda +13=0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle A=1\text{, } B=4 \text{ , } C=13 $$
 * $$\displaystyle A=1\text{, } B=4 \text{ , } C=13 $$


 * style= |
 * }

Quadratic formula:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \frac{-B\pm \sqrt{B^2-4AC}}{2A}=\frac{-4\pm \sqrt{16-52}}{2}=-2\pm \frac{\sqrt{-36}}{2}=-2\pm 3i $$
 * $$\displaystyle \frac{-B\pm \sqrt{B^2-4AC}}{2A}=\frac{-4\pm \sqrt{16-52}}{2}=-2\pm \frac{\sqrt{-36}}{2}=-2\pm 3i $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda _{1} =-2+3i \text{, } \lambda _{2}=-2-3i $$
 * $$\displaystyle \lambda _{1} =-2+3i \text{, } \lambda _{2}=-2-3i $$


 * style= |
 * }

General solution:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_{h}=e^{-ax/2}[A\cos\omega x+B\sin\omega x]  $$
 * $$\displaystyle y_{h}=e^{-ax/2}[A\cos\omega x+B\sin\omega x]  $$


 * style= |
 * }

Where,


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda=-\frac{1}{2}a\pm i\omega  $$
 * $$\displaystyle \lambda=-\frac{1}{2}a\pm i\omega  $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_{h}=e^{-2x}[A\cos3x+B\sin3x]  $$
 * $$\displaystyle y_{h}=e^{-2x}[A\cos3x+B\sin3x]  $$


 * style= |
 * }

Using initial conditions:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y(0)=1=e^0[A\cos0+B\sin0]=1[A(1)+B(0)]=A $$
 * $$\displaystyle y(0)=1=e^0[A\cos0+B\sin0]=1[A(1)+B(0)]=A $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'_{h}=e^{-2x}[-3A\sin3x+3B\cos3x]-2e^{-2x}[A\cos3x+B\sin3x]  $$
 * $$\displaystyle y'_{h}=e^{-2x}[-3A\sin3x+3B\cos3x]-2e^{-2x}[A\cos3x+B\sin3x]  $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'(0)=0=e^{0}[-3(1)\sin0+3B\cos0]-2e^{0}[(1)\cos0+B\sin0]=3B-2\therefore B=\frac{2}{3}   $$
 * $$\displaystyle y'(0)=0=e^{0}[-3(1)\sin0+3B\cos0]-2e^{0}[(1)\cos0+B\sin0]=3B-2\therefore B=\frac{2}{3}   $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.1.9)
 * }

Shown below is a plot of the solution.




 * {| style="width:100%" border="0"


 * style="width:100%" |
 * style="width:100%" |

Shown below is a superimposed graph of solutions to this problem (blue), R 2.1 (red), and R 2.6 (green).


 * style= |
 * }



Author
Solved and typed by - Silvio Attanasio,Nate Wanzie Reviewed By - Entire Group

=Contributing Members=