User:Egm4313.s12.team20/R2

Given
For scenario a) and b), find the differential equation with the given information. There is no excitation $$\displaystyle [r(x)=0]$$,so the differential equation becomes homogeneous. a)$$\displaystyle \lambda_1=3, \lambda_2=-6, y(0)=-5, y'(0)=2$$ b)$$\displaystyle \lambda_1=-2, \lambda_2=5, y(0)=1, y'(0)=0$$ .

Solution Part A
The characteristic equation for non-repeating roots is:
 * {| style="width:100%" border="0"|-

$$ \displaystyle y(x)=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}$$ When $$\displaystyle\lambda_1$$ & $$\displaystyle\lambda_2$$ are substituted into the characteristic equation, the following equation is found:
 * style="width:95%" |
 * (1.0)
 * }
 * {| style="width:100%" border="0"|-

$$ \displaystyle y(x)=C_1e^{3x}+C_2e^{-6x}$$ To be able to solve with both initial conditions, $$\displaystyle y'(x)$$ must be evaluated. The derivative is easily found to be:
 * style="width:95%" |
 * (1.1)
 * }
 * {| style="width:100%" border="0"|-

$$ \displaystyle y'(x)=3C_1e^{3x}-6C_2e^{-6x}$$ After applying the initial conditions, the following set of equations are obtained. For $$\displaystyle y(x)$$'s initial conditions:
 * style="width:95%" |
 * (1.2)
 * }
 * {| style="width:100%" border="0"|-

$$ \displaystyle -5=C_1e^{0}+C_2e^{0}$$ Which reduces to:
 * style="width:95%" |
 * (1.3)
 * }
 * {| style="width:100%" border="0"|-

$$ \displaystyle -5=C_1+C_2$$
 * style="width:95%" |
 * (1.4)
 * }

For $$\displaystyle y'(x)$$'s initial conditions:
 * {| style="width:100%" border="0"|-

$$ \displaystyle 2=3C_1e^{0}-6C_2e^{0}$$ Which reduces to:
 * style="width:95%" |
 * (1.5)
 * }
 * {| style="width:100%" border="0"|-

$$ \displaystyle 2=3C_1-6C_2$$ After solving equation (1.4) and equation (1.6) as a system of equations: $$\displaystyle \rightarrow C_1=-\frac{28}{9}$$ $$\displaystyle \rightarrow C_2=-\frac{17}{9}$$ The solution for the differential equation becomes:
 * style="width:95%" |
 * (1.6)
 * }

Solution Part B
The characteristic equation for non-repeating roots is:
 * {| style="width:100%" border="0"|-

$$ \displaystyle y(x)=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}$$ When $$\displaystyle\lambda_1$$ & $$\displaystyle\lambda_2$$ are substituted into the characteristic equation, the following equation is found:
 * style="width:95%" |
 * (1.8)
 * }
 * {| style="width:100%" border="0"|-

$$ \displaystyle y(x)=C_1e^{-2x}+C_2e^{5x}$$ To be able to solve with both initial conditions, $$\displaystyle y'(x)$$ must be evaluated. The derivative is easily found to be:
 * style="width:95%" |
 * (1.9)
 * }
 * {| style="width:100%" border="0"|-

$$ \displaystyle y'(x)=-2C_1e^{-2x}+5C_2e^{5x}$$ After applying the initial conditions, the following set of equations are obtained. For $$\displaystyle y(x)$$'s initial conditions:
 * style="width:95%" |
 * (1.10)
 * }
 * {| style="width:100%" border="0"|-

$$ \displaystyle 1=C_1e^{0}+C_2e^{0}$$ Which reduces to:
 * style="width:95%" |
 * (1.11)
 * }
 * {| style="width:100%" border="0"|-

$$ \displaystyle 1=C_1+C_2$$
 * style="width:95%" |
 * (1.12)
 * }

For $$\displaystyle y'(x)$$'s initial conditions:
 * {| style="width:100%" border="0"|-

$$ \displaystyle 0=-2C_1e^{0}+5C_2e^{0}$$ Which reduces to:
 * style="width:95%" |
 * (1.13)
 * }
 * {| style="width:100%" border="0"|-

$$ \displaystyle 0=-2C_1+5C_2$$ After solving equation (1.12) and equation (1.14) as a system of equations: $$\displaystyle \rightarrow C_1=\frac{5}{7}$$ $$\displaystyle \rightarrow C_2=\frac{2}{7}$$ The solution for the differential equation becomes:
 * style="width:95%" |
 * (1.14)
 * }

R 2.2
Find and plot the solution to Equation 2.0 below

Given
Second Order Linear ODE with Constant Coefficients:
 * {| style="width:100%" border="0"

$$\displaystyle y''-10y'+25y=r(x)$$ (2.0) Initial Conditons:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y(0)=1$$ (2.1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y''(0)=0$$ (2.2) No excitation:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle r(x)=0$$ (2.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
Since there is no excitation (r(x)=0), the ODE now becomes: The characteristic equation of Equation 2.4 is: Solve for the zeros of the characteristic equation:
 * {| style="width:100%" border="0"|-

$$\displaystyle \lambda^2-10\lambda+25=0$$ $$\displaystyle (\lambda-5)^2=0$$
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"|-
 * {| style="width:100%" border="0"|-

$$\displaystyle \lambda=5, 5$$ Since this is a real double root, the general solution of Equation 2.4 is:
 * style="width:95%" |
 * <p style="text-align:right">(2.6)
 * }
 * {| style="width:100%" border="0"|-

$$\displaystyle y=(c_{1} + c_{2}x)e^{-\lambda x}$$ Substitute Lambda:
 * style="width:95%" |
 * }
 * }
 * {| style="width:100%" border="0"|-

$$\displaystyle y=(c_{1} + c_{2}x)e^{-5x}$$ The particular solution is found by solving for constants c1 and c2 Initial condition 1 (Equation 2.1) is applied to Equation 2.7 to find c1:
 * style="width:95%" |
 * <p style="text-align:right">(2.7)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y(0)=1$$ $$\Downarrow$$ $$\displaystyle (c_{1} + c_{2}*0)e^{-5*0}=1$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle c_{1}=1$$ (2.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Derive Equation 2.7 using the Product Rule:
 * {| style="width:100%" border="0"

$$\displaystyle y=(c_{1} + c_{2}x)e^{-5x}$$ $$\displaystyle y=c_{1}e^{-5x} + c_{2}xe^{-5x}$$ $$\Downarrow$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle y'=-5c_{1}e^{-5x} + c_{2}e^{-5x} - 5c_{2}xe^{-5x}$$ (2.9) Initial condition 2 (Equation 2.2) and c1 (Equation 2.8) is applied to Equation 2.9 to find c2:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y'(0)=0 \, \ c_{1}=1$$ $$\Downarrow$$ $$\displaystyle 0=-5*1*e^{-5*0} + c_{2}e^{-5*0} - 5c_{2}*0*e^{-5*0}$$ $$\displaystyle 0=-5+c_{2}$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle c_{2}=5$$ (2.10) Substitute Equations 2.8 and 2.10 into 2.7 to find the particular solution:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"|-

$$\displaystyle y=(c_{1} + c_{2}x)e^{-5x} \, \ c_{1}=1 \ , \ c_{2}=5$$ $$ \Downarrow $$ Plot of 2.11:
 * style="width:95%" |
 * }
 * }

R 2.3
Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59 problems.3-4

Problem 3
Find a general solution. Check your answer by substitution.

Given

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

$$\displaystyle y''+6y'+8.96y = 0$$ (3.0)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$\frac{d^2 y}{dx^2} + 6\frac{dy}{dx} + 8.96y=0 $$ (3.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Set value for lambda
 * {| style="width:100%" border="0"

$$\lambda = \frac{dy}{dx}$$ (3.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Plug in the variable lambda to get the characteristic equation
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+6\lambda+8.96=0$$ (3.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Using the quadratic formula, the Distinct roots are determined to be
 * {| style="width:100%" border="0"

$$\displaystyle \lambda =-2.8,-3.2$$ (3.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The General Solution formula for Complex Conjugate roots is
 * {| style="width:100%" border="0"

$$\displaystyle y(x)=c_1 e^{\lambda_1 x}+c_2 e^{\lambda_2 x}$$ (3.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

Problem 4
Find a general solution. Check your answer by substitution.

Given

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

$$\displaystyle y''+4y'+(\pi^2 + 4)y = 0$$ (3.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
Rewrite as


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

$$\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y=0 $$ (3.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Set value for lambda
 * {| style="width:100%" border="0"

$$\lambda = \frac{dy}{dx}$$ (3.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Plug in the variable lambda to get the characteristic equation
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+4\lambda+(\pi^2+4)=0$$ (3.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The formula for the Complex Conjugate roots is
 * {| style="width:100%" border="0"

$$\lambda=-\frac{1}{2}\alpha \pm \beta i$$ (3.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Using the quadratic formula, the roots are determined to be
 * {| style="width:100%" border="0"

$$\lambda =-2\pm (\pi)i$$ (3.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The General Solution formula for Complex Conjugate roots is
 * {| style="width:100%" border="0"

$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1\cos \beta x+c_2\sin \beta x)$$ (3.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

R 2.4
Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59 problems.5-6

Problem 5
Find a general solution. Check your answer by substitution.

Given

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

$$\displaystyle y''+2\pi y'+\pi^2y = 0$$ (4.0)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
Rewrite as


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

$$\frac{d^2 y}{dx^2} + 2\pi \frac{dy}{dx} + (\pi^2 y)=0 $$ (4.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Set value for lambda
 * {| style="width:100%" border="0"

$$\lambda = \frac{dy}{dx}$$ (4.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Plug in the variable lambda to get the characteristic equation
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+2\pi \lambda+\pi^2 y=0$$ (4.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The formula for the Real Double root is
 * {| style="width:100%" border="0"

$$\lambda=-\frac{1}{2}\alpha$$ (4.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Using the quadratic formula, the root is determined to be
 * {| style="width:100%" border="0"

$$\displaystyle\lambda=-\pi$$ (4.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The General Solution formula for Real Double root is
 * {| style="width:100%" border="0"

$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1 + c_2 x)$$ (4.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

Problem 6
Find a general solution. Check your answer by substitution.

Given

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

$$\displaystyle 10y''-32 y'+25.6y = 0$$ (4.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
Rewrite as


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

$$10\frac{d^2 y}{dx^2} + 32 \frac{dy}{dx} + 25.6y=0 $$ (4.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Set value for lambda
 * {| style="width:100%" border="0"

$$\lambda = \frac{dy}{dx}$$ (4.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Plug in the variable lambda to get the characteristic equation
 * {| style="width:100%" border="0"

$$\displaystyle 10\lambda^2+32 \lambda+25.6 y=0$$ (4.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The formula for the Real Double root is
 * {| style="width:100%" border="0"

$$\lambda=-\frac{1}{2}\alpha$$ (4.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Using the quadratic formula, the root is determined to be
 * {| style="width:100%" border="0"

$$\displaystyle\lambda=\frac{8}{5}$$ (4.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The General Solution formula for Real Double root is
 * {| style="width:100%" border="0"

$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1 + c_2 x)$$ (4.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

Statement
Find an ODE in the form $$\displaystyle y''-ay'+by=0   $$ for the given basis.

(16.) $$\displaystyle e^{2.6x} $$ , $$\displaystyle e^{-4.3x} $$

(17.) $$\displaystyle e^{-\sqrt 5 x}$$ , $$\displaystyle xe^{-\sqrt 5 x} $$

Background information
We will consider second-order homogeneous linear ODEs with constant coefficients.

$$\displaystyle y''=ay'=by=0 $$

Knowing that the solution to a first order linear ODE with a constant coefficient k

$$\displaystyle y'+ky=0 $$

is an exponential function

$$\displaystyle y=e^{\lambda x} $$

Substituting it and its derivatives

$$\displaystyle y'=\lambda e^{\lambda x} $$

$$\displaystyle y''=\lambda^2 e^{\lambda x}$$

into $$\displaystyle y''-ay'+by=0   $$

This results in

$$\displaystyle \lambda^2 + a \lambda + b = 0 $$

Problem 16
Using the following basis

$$\displaystyle e^{2.6x}$$

$$\displaystyle e^{-4.3x}$$

The general solution is

$$\displaystyle y=c_{1} e^{\lambda_{1} x} + c_{2} e^{\lambda_{2} x} $$

Plugging in the basis given in (16.) into the above equation we get

$$\displaystyle y=c_{1} e^{2.6x} + c_{2} e^{-4.3x} $$

From this we see that

$$\displaystyle \lambda_{1} = 2.6 $$

$$\displaystyle \lambda_{2} = -4.3 $$

Using properties of quadratic equations,

$$\displaystyle (\lambda - 2.6)(\lambda + 4.3) = 0 $$

$$\displaystyle \lambda^2 + 4.3 \lambda - 2.6 \lambda -11.18 = 0$$

$$\displaystyle \lambda^2 + 1.7\lambda - 11.18 = 0 $$

Because of the relationship shown boxed above in the background information, this results in

$$\displaystyle y'' + 1.7y'-11.18y=0 $$

Problem 17
Using the following basis

$$\displaystyle e^{-\sqrt5 x} $$

$$\displaystyle e^{-\sqrt5 x} $$

The general solution is $$\displaystyle y=c_{1} e^{\lambda_{1} x} + c_{2} e^{\lambda_{2} x} $$

Plugging in the basis into the above equation we get

$$\displaystyle y=c_{1} e^{-\sqrt5 x} + c_{2} e^{-\sqrt5 x} $$

Where

$$\displaystyle \lambda_{1} = \lambda_{2} $$

Using properties of quadratic equations, we solve

$$\displaystyle (\lambda + \sqrt5)(\lambda + \sqrt5) = 0 $$

$$\displaystyle \lambda^2 + \sqrt5 \lambda +\sqrt5 \lambda +5 = 0$$

$$\displaystyle \lambda^2 + 2 \sqrt5 \lambda + 5 = 0 $$

Because of the relationship shown boxed above in the background information, this results in




 * $$\displaystyle y'' + 2\sqrt5 y'+5y=0 $$
 * }

Statement
Realize a spring-dashpot-mass system in series as shown below in the figure. Consider the double real root $$\displaystyle \lambda = -3 $$ and find the values for the parameters k,c,m.

Figure



Solution
The equation of motion for a spring dashpot-mass system in series is


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

$$f(t) = m(y''_k + \frac{k}{c}y'_k)+ky_k $$     (6.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Dividing through by m, we obtain


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

$$f(t) = y''_k + \frac{k}{cm}y'_k+ \frac{k}{m}y_k $$     (6.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Recall, the original homogeneous L2-ODE-CC is


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

$$\displaystyle y''_h +ay'_h+by_h = 0 $$     (6.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, by relating equation (6.3) to equation (6.2), we obtain


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

$$a = \frac{k}{cm} $$     (6.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$b= \frac{k}{m} $$     (6.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

With $$\displaystyle \lambda = -3 $$, the characteristic equation is


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

$$\displaystyle (\lambda+3)^2 \rightarrow \lambda^2+6\lambda+9=0 $$     (6.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The general solution $$\displaystyle (\because\lambda=y')$$ is
 * {| style="width:100%" border="0"

$$\displaystyle y''+6y'+9y=0 $$     (6.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Relating equation of the general solution (6.8) to the original homogeneous L2-ODE-CC equation (6.3), we conclude


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

$$\displaystyle a=6 $$     (6.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle b=9 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

After plugging in the values of $$\displaystyle a and b $$ to equation (6.5) and (6.6), we obtain


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

$$\displaystyle 6=\frac{k}{cm} $$     (6.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle 9=\frac{k}{m} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, three possible values for $$\displaystyle k,c,m $$ are

Author:
Solved and typed by: Egm4313.sp12.team20.sablotsky.ca 20:20, 6 February 2012 (UTC)

Statement
Develop the MacLaurin series (Taylor series at t=0)for $$ \displaystyle e^t,cos(t), sin(t). $$

Solution
The general form for the taylor series expansion in sigma notation is


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

$$ \displaystyle \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (t-a)^n $$
 * style="width:95%" |
 * <p style="text-align:right">(7.0)
 * }

Writing the expansion out, we obtain


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

$$ \displaystyle f(a) + \frac{f'(a)}{1!}(t - a) + \frac{f(a)}{2!}(t - a)^2 + \frac{f'(a)}{3!}(t - a)^3 + ... $$
 * style="width:95%" |
 * <p style="text-align:right">(7.1)
 * }

A.

The Maclaurin Series of $$\displaystyle e^t $$ at t=0 is


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

$$ \displaystyle e^t = \sum_{n=0}^{\infty} \frac{t^n}{n!} $$
 * style="width:95%" |
 * <p style="text-align:right">(7.2)
 * }

After expanding this out, we obtain that $$\displaystyle e^t $$ equals


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

$$ \displaystyle e^t = e^0 + \frac{e^0}{1!}(t) + \frac{e^0}{2!}(t)^2 + \frac{e^0}{3!}(t)^3 + ... $$
 * style="width:95%" |
 * <p style="text-align:right">(7.3)
 * }

With $$\displaystyle e^0 = 1 $$ the final solution is

B.

The Maclaurin Series of $$\displaystyle cos(t) $$ at t=0 is


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

$$ \displaystyle cos(t) = \sum_{n=0}^{\infty} \frac{(-1)^nt^{2n}}{(2n)!} $$
 * style="width:95%" |
 * <p style="text-align:right">(7.5)
 * }

After expanding this out, we obtain


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

$$ \displaystyle cos(0) + \frac{-sin(0)}{1!}(t) + \frac{-cos(0)}{2!}(t)^2 + \frac{sin(0)}{3!}(t)^3 + \frac{cos(0)}{4!}(t)^4 + ... $$
 * style="width:95%" |
 * <p style="text-align:right">(7.6)
 * }

With $$\displaystyle cos(0) = 1 $$ the final solution is

C.

The Maclaurin Series of $$\displaystyle sin(t) $$ at t=0 is


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

$$ \displaystyle sin(t) = \sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(0)(t)^n \! $$
 * style="width:95%" |
 * <p style="text-align:right">(7.8)
 * }

After expanding this out, we obtain


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

$$ \displaystyle sin(0) + \frac{cos(0)}{1!}(t) + \frac{-sin(0)}{2!}(t)^2 + \frac{-cos(0)}{3!}(t)^3 + \frac{sin(0)}{4!}(t)^4 + ... $$
 * style="width:95%" |
 * <p style="text-align:right">(7.9)
 * }

With $$\displaystyle sin(0) = 0 $$ the final solution is

Author:
Solved and Typed By: Egm4313.sp12.team20.sablotsky.ca 20:21, 6 February 2012 (UTC)

Statement
Complete Kreyszig 2011 p.59 Problems: 8 & 15. <br \>

Given
Find a general solution. Check your answer by substitution: <br \>


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

$$  \displaystyle y'' + y' + 3.25y = 0 $$     (8.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$     (8.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution
To find the corresponding general solution for each equation, the first step is to take each equation in its standard form, and generate a characteristic equation using $$ \lambda $$. The characteristic equation should be in the form of $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$. The next step is to find the roots to the corresponding characteristic equation. Lastly, use the roots to generate the necessary general solution with constants. After finding this general solution, we check it by taking the corresponding derivatives of the solution, then substituting it back into the original equation to see if everything cancels.

8.8


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

$$  \displaystyle y'' + y' + 3.25y = 0 $$     (8.8.0)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The characteristic equation of Eqn. 8.8.0 is:


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

$$  \displaystyle \lambda^2 + \lambda + 3.25 = 0 $$     (8.8.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

To find the roots of Eqn. 8.8.1, we use the quadratic formula:


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

$$  \displaystyle \lambda_1,\lambda_2 = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{-1 \pm \sqrt{1^2-4(1)(3.25)}}{2(1)} $$     (8.8.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Eqn. 8.8.2 results in the following complex roots, as a result of the "-" in the square-root:


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

$$  \displaystyle \lambda_1,\lambda_2 = \frac{-1}{2} \pm \sqrt{3}i $$     (8.8.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Then, we plug the previous complex roots, in the form $$ x = r \pm ti $$ into the following general solution equation:


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

$$  \displaystyle y = e^{rx}(Acos(tx)+Bsin(tx)) $$     (8.8.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

8.8 Check Derivatives:

$$\displaystyle y = e^{\frac{-x}{2}}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))$$ $$\displaystyle y' = e^{\frac{-x}{2}}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))-\frac{1}{2}e^{\frac{-x}{2}}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))$$ which simplifies to: $$\displaystyle y' = e^{\frac{-x}{2}}((-\sqrt{3}A-\frac{B}{2})sin(\sqrt{3}x)+(\sqrt{3}B-\frac{A}{2})cos(\sqrt{3}x))$$ $$\displaystyle y'' = e^{\frac{-x}{2}}(\sqrt{3}(-\sqrt{3}A-\frac{B}{2})cos(\sqrt{3}x)-\sqrt{3}(\sqrt{3}B-\frac{A}{2})sin(\sqrt{3}x))...$$ $$\displaystyle -\frac{1}{2}e^{\frac{-x}{2}}((-\sqrt{3}A-\frac{B}{2})sin(\sqrt{3}x)+(\sqrt{3}B-\frac{A}{2})cos(\sqrt{3}x))$$ which simplifies to: $$\displaystyle y'' = e^{\frac{-x}{2}}((-\sqrt{3}B-\frac{11}{4}A)cos(\sqrt{3}x)+(\sqrt{3}A-\frac{11}{4}B)sin(\sqrt{3}x))$$

Substitutions:

Plugging everything back into the original ODE: $$\displaystyle e^{\frac{-x}{2}}((-\sqrt{3}B-\frac{11}{4}A)cos(\sqrt{3}x)+(\sqrt{3}A-\frac{11}{4}B)sin(\sqrt{3}x))...$$ $$\displaystyle +e^{\frac{-x}{2}}((-\sqrt{3}A-\frac{B}{2})sin(\sqrt{3}x)+(\sqrt{3}B-\frac{A}{2})cos(\sqrt{3}x))...$$ $$\displaystyle +(3.25)[e^{\frac{-x}{2}}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))]=0$$ The $$e^{\frac{-x}{2}}$$ can be cancelled out and with some manipulation, you get: $$\displaystyle (-\frac{11}{4}A-\sqrt{3}B+\sqrt{3}B-\frac{A}{2}+3.25A)cos(\sqrt{3}x)+(-\frac{11}{4}B-\sqrt{3}A+\sqrt{3}A-\frac{B}{2}+3.25B)sin(\sqrt{3}x)=0$$ Summing all the values in each parenthesis, we find that the value becomes 0, so everything on the left side becomes 0: $$\displaystyle (0)cos(\sqrt{3}x)+(0)sin(\sqrt{3}x)=0=0$$ So the general solution found in the box above is in fact a true general solution to the original ODE based on this check

8.15


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

$$  \displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$     (8.15.0)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The characteristic equation of Eqn. 8.15.0 is:


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

$$  \displaystyle \lambda^2 + 0.54\lambda + (0.0729 + \pi) = 0 $$     (8.15.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

To find the roots of Eqn. 8.15.1, we use the quadratic formula:


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

$$  \displaystyle \lambda_1,\lambda_2 = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{-0.54 \pm \sqrt{0.54^2-4(1)(0.0729 + \pi)}}{2(1)} $$     (8.15.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Eqn. 8.15.2 results in the following complex roots, as a result of the "-" in the square-root:


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

$$  \displaystyle \lambda_1,\lambda_2 = -0.27 \pm \sqrt{\pi}i $$     (8.15.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Then, we plug the previous complex roots, in the form $$ x = r \pm ti $$ into the following general solution equation:


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

$$  \displaystyle y = e^{rx}(Acos(tx)+Bsin(tx)) $$     (8.15.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

8.15 Check Derivatives:

$$\displaystyle y = e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))$$ $$\displaystyle y' = e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))-0.27e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))$$ which simplifies to: $$\displaystyle y' = e^{-0.27x}((-\sqrt{\pi}A-0.27B)sin(\sqrt{\pi}x)+(\sqrt{\pi}B-0.27A)cos(\sqrt{\pi}x))$$ $$\displaystyle y'' = e^{-0.27x}(\sqrt{\pi}(-\sqrt{\pi}A-0.27B)cos(\sqrt{\pi}x)-\sqrt{\pi}(\sqrt{\pi}B-0.27A)sin(\sqrt{\pi}x))...$$ $$\displaystyle -0.27e^{-0.27x}((-\sqrt{\pi}A-0.27B)sin(\sqrt{\pi}x)+(\sqrt{\pi}B-0.27A)cos(\sqrt{\pi}x))$$ which simplifies to: $$\displaystyle y'' = e^{-0.27x}(((-\pi+0.27^2)A-0.54\sqrt{\pi}B)cos(\sqrt{\pi}x)+((-\pi+0.27^2)B+0.54\sqrt{\pi}A)sin(\sqrt{\pi}x))$$

Substitutions:

Plugging everything back into the original ODE: $$\displaystyle e^{-0.27x}(((-\pi+0.27^2)A-0.54\sqrt{\pi}B)cos(\sqrt{\pi}x)+((-\pi+0.27^2)B+0.54\sqrt{\pi}A)sin(\sqrt{\pi}x))...$$ $$\displaystyle +(0.54)[e^{-0.27x}((-\sqrt{\pi}A-0.27B)sin(\sqrt{\pi}x)+(\sqrt{\pi}B-0.27A)cos(\sqrt{\pi}x))]...$$ $$\displaystyle +(0.0729+\pi)[e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))]=0$$ The $$e^{-0.27x}$$ can be cancelled out and with some manipulation, you get: $$\displaystyle ((-\pi+0.27^2)A-0.54\sqrt{\pi}B+0.54\sqrt{\pi}B-(0.54)(0.27)A+0.0729A+\pi A)cos(\sqrt{\pi}x)...$$ $$\displaystyle +((-\pi+0.27^2)B-0.54\sqrt{\pi}A+0.54\sqrt{\pi}A-(0.54)(0.27)B+0.0729B+\pi B)sin(\sqrt{\pi}x)=0$$ Summing all the values in each parenthesis, we find that the value becomes 0, so everything on the left side becomes 0: $$\displaystyle (0)cos(\sqrt{\pi}x)+(0)sin(\sqrt{\pi}x)=0=0$$ So the general solution found in the box above is in fact a true general solution to the original ODE based on this check

Author:
Solved and Typed By: --Egm4313.s12.team20.calvo 18:45, 8 February 2012 (UTC)

Statement
Find and plot the solution to the L2-ODE-CC corresponding to $$ \displaystyle  \lambda_1 = -2 + 3i $$ and $$ \displaystyle \lambda_2 = -2 - 3i $$ and the initial conditions $$ \displaystyle y(0) = 1, y'(0) = 0 $$ with an excitation of $$ \displaystyle r(x) = 0 $$

Solution
Given $$ \displaystyle \lambda_1 $$ and $$ \displaystyle \lambda_2 $$, our characteristic equation is:

<p style="text-align:left"> Noting
 * {| style="width:100%" border="0"

$$\displaystyle y'' + 4y' + 13y = r(x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(2.9.1)
 * }

<p style="text-align:left"> Since r(x) = 0, this just becomes the homogeneous equation

<p style="text-align:left"> Noting that {| style="width:100%" border="0"|- $$ \displaystyle w^2 = b - (1/4)a^2 $$
 * style="width:95%" |

<p style="text-align:left"> We get a general solution of

<p style="text-align:left"> From the initial conditions we obtain {| style="width:100%" border="0"|- $$ \displaystyle y(0) = 1 = A $$ {| style="width:100%" border="0"|- $$ \displaystyle y'(0) = 0 = 1(0 + 3B) - 2(A) $$ {| style="width:100%" border="0"|- $$ \displaystyle B = 2/3 $$
 * style="width:95%" |
 * style="width:95%" |
 * style="width:95%" |

<p style="text-align:left"> Thus, our final solution is

<p style="text-align:left">The plot of the final solution: <p style="text-align:left">x = linspace(0,10); <p style="text-align:left">y = exp(-2*x).*(cos(3*x)+((2/3)*sin(3*x))); <p style="text-align:left">plot(x,y)

<p style="text-align:left"> The superpose of the above plot and plots from R2.6 and R2.1: <p style="text-align:left">x = linspace(0,10); <p style="text-align:left">y = exp(-2*x).*(cos(3*x)+((2/3)*sin(3*x))); <p style="text-align:left">y1 = (1+(5*x)).*exp(-5*x); <p style="text-align:left">y2 = exp(-2*x).*(cos(3*x)+((2/3)*sin(3*x))); <p style="text-align:left">plot(x,y,x,y1,x,y2) <p style="text-align:left">(NOTE: function y2 does not fit on the graph in a way that the other two can still be seen)

<p style="text-align:left">===Author:=== <p style="text-align:left">Solved and Typed by: Egm4313.sp12.team20.samartino 17:47, 6 February 2012 (UTC)