User:Egm4313.s12.team2/Report2

=R2.1=

Problem Statement
Given two roots and the initial conditions:
 * $$ \lambda _1 = -2 \ $$, $$ \lambda _2 = +5 \ $$
 * $$ y(0) = 1 \ $$, $$ y'(0) = 0 \ $$

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation r(x). Consider $$ r(x) = 0 \ $$. Plot the solution.

Soluion-Part 1
Given:
 * $$ \lambda_1 = -2, \lambda_2 = 5 \ $$
 * $$ y(0) = 1, y'(0) = 0 \ $$

With these roots, the original,standard ODE can be found by finding the characteristic equation.
 * $$ (\lambda - (-2))(\lambda - 5) = 0 \ $$
 * $$ \lambda^2 - 3\lambda - 10 = 0 \ $$

The solution is the sum of the homogenous solution and the particular solution. Since r(x) is unknown, yp will represent the particular solution.
 * $$ y(t) = y_h + y_p \ $$

The roots are real and distinct so the homogenous solution takes the form:
 * $$ y_h = C_1 e^{-2t} + C_2 e^{5t} \ $$

By taking the derivative and substituting the initial conditions, the following system is derived to calculate the coefficients.
 * $$ y(0) = 1 = C_1 + C_2 \ $$
 * $$ y'(0) = 0 -2C_1 + 5C_2 \ $$
 * $$ C_1 = 1- C_2 \ $$
 * $$ 0 = -2 + 2C_2 + 5C_2 \ $$
 * $$ 2 = 7C_2 \ $$
 * $$C_1 = \frac{5}{7}, C_2 = \frac{2}{7} \ $$

The solution to the nonhomogenous L2-ODE is
 * $$ y(t) = \frac{5}{7} e^{-2t} + \frac{2}{7} e^{5t} + y_p \ $$

If r(x)=0, Yp=0 then y(t)= yh. The plot is shown below.



Solution-Part 2
Given:
 * $$ \lambda_1 = 3,\lambda_2 = -6 \ $$

The characteristic equation is
 * $$ (\lambda - 3)(\lambda - (-6)) = 0 \rightarrow \lambda^2 + 3\lambda -18 = 0 \ $$

The standard L2-ODE is
 * $$ y'' + 3y' - 18y = r(x) \ $$

Non-standard
 * $$ m(y'' + 3y' -18y) = r(x) \ $$

3 non-standard equations can be found by multiplying the standard form by a constant,m. Choosing m=3,5,7 respectively.
 * $$ 3y'' + 9y' - 54y = r(x) \ $$
 * $$ 5y'' + 15y' - 90y = r(x) \ $$
 * $$ 7y'' + 21y' - 126y = r(x) \ $$

--John Plunkett 17:30, 7 February 2012 (UTC)

=R2.2=

Problem Statement
Given initial conditions $$ y(0) = 1 \ $$, $$ y'(0) = 0 \ $$ and excitation $$ r(x) = 0 \ $$ , find the solution to $$ y'' - 10y' + 25y = r(x) \ $$

Solution

 * $$ y'' - 10y' + 25y = r(x) \ $$
 * $$ \lambda^2 - 10\lambda + 25 = 0 \ $$
 * $$ (\lambda - 5)(\lambda - 5) = 0 \ $$
 * $$ \lambda = 5 \ $$

Double real root
 * $$ \lambda = -\frac 1 2 a $$
 * $$ y = (C_1 + C_2 x)e^{- \frac {ax} 2} \ $$
 * $$ y = C_1 e^{5x} + C_2 xe^{5x} \ $$

Apply initial conditions
 * $$ 1 = C_1 e^{5(0)} + C_2 (0)e^{5(0)} \ $$
 * $$ 1 = C_1 e \ $$
 * $$ C_1 = 1 \ $$
 * $$ y' = 5C_1 e^{5x} + C_2 (1*e^{5x}) + x*5xe^{5x} \ $$
 * $$ y' = 5C_1 e^{5x} + C_2 e^{5x} + 5x^2 e^{5x} \ $$
 * $$ y' = 5(1) e^{5(0)x} + C_2 e^{5(0)} + 5(0)^2 e^{5(0)} \ $$
 * $$ 0 = 5+ C_2 + 0 \ $$
 * $$ C_2 = -5 \ $$
 * $$ y = (1) e^{5x} + (-5) xe^{5x} \ $$

$$ y = e^{5x}-5xe^{5x} \ $$ --Corey Williams 17:55, 6 February 2012 (UTC)

Matlab Code
x = 0:0.01:1; >> ft = exp(5*x) - 5*x.*exp(5*x); >> plot(x,ft)

Plot
--Corey Williams 15:10, 7 February 2012 (UTC)

=R2.3=

Problem Statement
K 2011 p59 #3 and 4: Find a general solution, check by substitution.

Problem 3

 * $$ y'' + 6y' + 8.96y = 0 \ $$
 * $$ r^2 + 6r + 8.96 = 0 \ $$
 * $$ r = (\frac{-6 \pm \sqrt{36 - 4(8.96)}}{2}) \ $$
 * $$ r = -3 \pm 0.2 \ $$
 * $$ r = -2.8, -3.2 $$

Therefore, the general solution takes the form of: $$ y_g = C_1 e^{-2.8x} + C_2 e^{-3.2x} \ $$
 * $$ y' = -2.8C_1 e^{-2.8x} - 3.2C_2 e^{-3.2x} \ $$


 * $$ y'' = 7.84C_1 e^{-2.8x} + 10.24C_2 e^{-3.2x} \ $$

Substitution back into the original equation should result in 0 = 0
 * $$ 0 = (7.84C_1 e^{-2.8x} + 10.24C_2 e^{-3.2x}) + 6 (-2.8C_1 e^{-2.8x} - 3.2C_2 e^{-3.2x}) + 8.96 (C_1 e^{-2.8x} + C_2 e^{-3.2x}) \ $$







Problem 4

 * $$ y'' + 4y' +(\pi^2 + 4)y = 0 \ $$
 * $$ r^2 + 4r = (\pi^2 + 4) = 0 \ $$
 * $$ r = (\frac{-4 \pm \sqrt{16 - 4(\pi^2 + 4)}}{2}) \ $$
 * $$ r = -2\pm \sqrt{4 - (\pi^2 + 4)} \ $$
 * $$ r = -2\pm\pi i \ $$

Therefore, the general solution takes the form of: $$ y_g = Ae^{-2t}cos(\pi t) + Be^{-2t}sin(\pi t) \ $$
 * $$ y' = A[-2e^{-2t}cos(\pi t) - \pi e^{-2t}sin(\pi t)] + B[-2e^{-2t}sin(\pi t) + \pi e^{-2t}cos(\pi t)] \ $$


 * $$ y'' = A[4e^{-2t}cos(\pi t) + 2\pi e^{-2t}sin(\pi t) + 2\pi e^{-2t}sin(\pi t) - \pi^2 e^{-2t}cos(\pi t)] \ $$


 * $$ + B[4e^{-2t}sin(\pi t) - 2\pi e^{-2t}cos(\pi t) - 2\pi e^{-2t}cos(\pi t) - \pi^2 e^{-2t}sin(\pi t)] \ $$

Substitution back into the original equation should result in 0 = 0
 * $$ 0 = A[4e^{-2t}cos(\pi t) + 2\pi e^{-2t}sin(\pi t) + 2\pi e^{-2t}sin(\pi t) - \pi^2 e^{-2t}cos(\pi t)] \ $$


 * $$ + B[4e^{-2t}sin(\pi t) - 2\pi e^{-2t}cos(\pi t) - 2\pi e^{-2t}cos(\pi t) - \pi^2 e{^-2t}sin(\pi t)]  \ $$


 * $$ + (4A[-2e^{-2t}cos(\pi t) - \pi e^{-2t}sin(\pi t)] + B[-2e^{-2t}sin(\pi t) + \pi e^{-2t}cos(\pi t)] + (\pi^2+4)(Ae^{-2t}cos(\pi t) + Be^{-2t}sin(\pi t)) \ $$


 * $$ 0 = 0 \ $$

Substitution of the solution and its derivatives into the original differential equation returns a value of zero, thereby proving that the solution is correct.

=R2.4=

Problem Statement
K 2011 p59 #5 and 6: Find a general solution, check by substitution.

Problem 5
$$ y'' +2\pi y' +\pi y = 0 \ $$ $$ r^2 +2\pi r +\pi ^2 = 0 \ $$ $$r = \frac{-2\pi \pm \sqrt{4\pi ^2 - 4\pi^2} }{2} \ $$ $$ r= \frac{-2\pi}{2} = -\pi\ $$ Therefore, the general solution takes the form of:

$$ y_g = C_1e^{-\pi t} + C_2te^{-\pi t} \ $$ $$ y' = -\pi C_1e^{-\pi t} + C_2[e^{-\pi t} -\pi te^{-\pi t}] \ $$ $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi e^{-\pi t} - \pi(e^{-\pi t}-\pi te^{-\pi t})] \ $$ $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi ^2te^{-\pi t} - 2\pi e^{-\pi t}] \ $$

Substitution back into the original equation should result as 0=0

$$0 = \pi^2C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} -2\pi C_2e^{-\pi t} - 2\pi^2C_1e^{-\pi t} + 2\pi C_2e^{-\pi t} -2\pi^2C_2e^{-\pi t}+\pi^2 C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} \ $$ $$ 0 = 0 \ $$ Thus the substitution checks out, and the form of the general solution is proven.

--Stephen Leshko 20:10, 30 January 2012 (UTC)

Problem 6

 * $$ 10y'' -32y' +25.6y = 0 \ $$

Which can be written in standard form as:
 * $$ y'' -3.2y' +2.56y = 0 \ $$

The characteristic equation is then:
 * $$ \lambda ^2 - 3.2\lambda +2.56 = 0 \ $$
 * $$ \lambda_1,\lambda_2 = \frac{3.2 \pm \sqrt{(-3.2)^2 - 4(2.56)}}{2} \ $$

Therefore the roots $$ \lambda_1, \lambda_2 = 1.6 \ $$ are actually a double root at $$ 1.6 $$ The form for the solution of the differential equation with double-real roots is of the following:
 * $$ y_h(t) = C_1e^{1.6t} + C_2te^{1.6t} \ $$

And then in order to check the solution, find the first and second derivatives to plug back into the original equation and check their equality to zero.
 * $$ y'_h(t) = 1.6C_1 e^{1.6t} + C_2[e^{1.6t} +1.6te^{1.6t}] \ $$
 * $$ y''_h(t) = 2.56C_1e^{1.6t} + C_2[1.6e^{1.6t} + 1.6e^{1.6t} + 2.56te^{1.6t}] \ $$


 * $$ 10[1.6C_1e^{1.6t} +C_2[3.2e^{1.6t} +2.56te^{1.6t}]] - 32[1.6C_1e^{1.6t} +C_2[e^{1.6t} + 1.6te^{1.6t}]] + 25.6[C_1e^{1.6t} +C_2te^{1.6t}] = 0 \ $$


 * $$ 25.6C_1e^{1.6t} + 32C_2e^{1.6t} + 25.6C_2te^{1.6t} - 51.2C_1e^{1.6t} -32C_2e^{1.6t} - 51.2C_2te^{1.6t} + 25.6C_1e^{1.6t} + 25.6C_2te^{1.6t} = 0 \ $$

The solution then checks out

--Stephen Leshko 21:28, 5 February 2012 (UTC)

=R2.5=

Problem Statement
K 2011 p59 #16 and 17: Find an ODE $$ y'' +ay' + by = 0 \ $$ for the given basis.

Problem 16
Basis: $$ e^{2.6x}, e^{-4.3x} \ $$


 * $$ \lambda_1,\lambda_2 =2.6, -4.3 = \frac{-a \pm \sqrt{a^2 - 4b}}{2} \ $$

Therefore multiplying each $$ \lambda \ $$ by 2 yields:
 * $$ 5.2 = -a + \sqrt{a^2 - 4b} \ $$ ............. (1)
 * $$ -8.6 = -a - \sqrt{a^2 -4b} \ $$ ............. (2)

Adding the equations together gives:
 * $$ -3.4 = -2a \ $$ so $$ a = 1.7 \ $$

Plug $$ a \ $$ back into equation (1) to get:
 * $$ 5.2 = -1.7 + \sqrt{1.7^2 - 4b} \ $$

Solving for $$ b \ $$:
 * $$ b = \frac {(5.2 +1.7)^2 - 1.7^2}{-4} = -11.18 \ $$

So then the differential equation would be: $$ y'' +1.7y' - 11.18 y = 0 \ $$

Problem 17
Basis: $$ e^{-\sqrt{5}x} ,xe^{-\sqrt{5}x} \ $$

From the basis it can be seen that $$ -\sqrt{5} \ $$ is a double root of the characteristic equation
 * $$ \lambda_1, \lambda_2 = - \sqrt{5} \ $$

So then the characteristic equation must have root such that:
 * $$ (\lambda + \sqrt{5})^2 = 0 \ $$

Multiplying out gives:
 * $$ \lambda ^2 + 2 \sqrt{5} \lambda + 5 = 0 \ $$

Which corresponds to the differential equation: $$ y'' + \sqrt{20} y' + 5y = 0 \ $$ --Stephen Leshko 21:31, 5 February 2012 (UTC)

=R2.6=

Problem Statement
Realize spring-dashpot-mass systems in series, as shown in Figure 2.6, with a double root $$ \lambda = -3 $$. Find the values for the parameters k, c, m.

Figure 2.6

Solution
The equation of motion for spring-dashpot-mass system in series is: $$ m(y''_k+\frac{k}{c}y'_k)+ky_k=f(t)\ $$ Distribute m: $$ my''_k+\frac{mk}{c}y'_k+ky_k=f(t)\ $$ The standard L2-ODE-CC with excitation is: $$ Ay''+By'+Cy=r(x)\ $$ The homogeneous equation is: $$ Ay''_h+By'_h+Cy_h=0\ $$ Given the double root: $$ \lambda=-3\ $$

The characteristic equation is: $$ [\lambda-(-3)]^2=(\lambda+3)^2=\lambda^2+6\lambda+9=0\ $$ The homogeneous L2-ODE-CC for this characteristic equation is: $$ y''+6y'+9y=0\ $$ Compare coefficients with: $$ my''_k+\frac{mk}{c}y'_k+ky_k=f(t)\ $$ The result is: $$ m=1\ $$

$$ k=9\ $$

$$ 6=\frac{mk}{c}\ $$

$$ 6=\frac{mk}{c}\rightarrow c=\frac{mk}{6}=\frac{(1)(9)}{6}\ $$

$$ c=\frac{3}{2}\ $$

--Jonathon Ladin 17:15, 07 February 2012 (UTC)

=R2.7=

=R2.8=

Problem Statement
K2011 pg. 59 #5,13 Find the general solution and check by substitution.

Problem 5
Given the following L2-0DE:
 * $$ y'' + y' + 3.25y = 0 \ $$

The corresponding characteristic equation is
 * $$ \lambda^2 + \lambda +3.25 = 0\ $$

The roots are
 * $$ \lambda_1, \lambda_2 = \frac{-1 \pm \sqrt{1-4*(3.25)}}{2} = -\frac{1}{2} \pm j\sqrt{3} \ $$

These roots yield the following solution.
 * $$ y(t) = C_1 e^{-\frac{t}{2}}cos(\sqrt{3}t) + C_2 e^{-\frac{t}{2}}sin(\sqrt{3}t) \ $$

After taking the derivatives and substituting them into the differential equation, the coefficients of the sine and cosine terms sum to zero. Cosine terms:
 * $$ 0 = (3.25C_1 - .5C_1 + \sqrt{3} C_2 + .25C_1 - 3C_1 -.5\sqrt{3}C_2 -.5\sqrt{3}C_2) e^{-.5t}cos(\sqrt{3}t) \ $$
 * $$ 0 = (3.25C_2 - \sqrt{3}C_1 - .5C_2 + .5\sqrt{3}C_1 +.5\sqrt{3}C_1 + .25C_2 -3C_2) e^{-.5t}sin(\sqrt{3}t) \ $$

Problem 13
Given the following L2-ODE:
 * $$ y'' + .54y' + (\pi+.0729)y = 0 \ $$

The Characteristic equation is
 * $$ \lambda^2 +.54\lambda + (\pi +.0729) = 0 \ $$

The roots are
 * $$ \lambda_1, \lambda_2 = \frac{-.54 \pm \sqrt{(.54)^2 - 4(\pi + .0729)}}{2} = -.27 \pm j(1.772)  \ $$

The roots can be used to find the following general, homogenous solution
 * $$ y(t) = C_1 e^{-.27t} cos(1.772t) + C_2 e^{-.27t} sin(1.772t) \ $$

After taking the derivatives and substituting them into the differential equation, the coefficients of the sine and cosine terms sum to zero.

Cosine terms:
 * $$ 0 = (3.2145C_1 - .1458C_1 + .9569C_2 + .0729C_1 - 3.14C_1 - .47844C_2 - .47844C_2) e^{-.27t} cos(1.772t) \ $$

Sine terms:
 * $$ 0 = (3.2145C_2 - .95688C_1 - .1458C_2 + .47844C_1 + .47844C_1 + .0729C_2 - 3.14C_2) e^{-.27t} sin(1.772t) \ $$

--Egm4313.s12.team2.plunkett.jp|John Plunkett 19:06, 8 February 2012 (UTC)

=R2.9=

Problem Statement
Given the initial conditions: Y(0)=1, Y'(0)=0 and the excitation r(x)=0  Find and plot the solution to the differential equation corresponding to $$ \lambda ^2+ 4 \lambda + 13 = r(x) $$ 

Solution
The differential equation that corresponds to $$ \lambda ^2 + 4 \lambda +13 = r(x) $$ is $$ y'' + 4y' + 13y = r(x) $$  Recognizing that the equation is equal to the excitation r(x) which was declared to be zero in the problem statement we can proceed to find the roots of the equation via the quadratic formula  $$ \lambda_1,\lambda_2 = \frac{-4 \pm \sqrt{(4)^2 - 4(13)(1)}}{2} \ = -2 \pm \frac{ /sqrt{16-52} } {2}$$ $$ \lambda_1,\lambda_2 = -2 \pm 3i $$ 

The general solution to a homogenous ODE with the complex roots $$ r_1, r_2=\alpha \pm \beta i$$takes on the general form:  $$ y(t)=C_1 e ^{\alpha t} cos(\beta t) + C_2 e ^{\alpha t} sin(\beta t) \ $$ 

Given $$ \lambda_1,\lambda_2 = -2 \pm 3i $$ we can substitute $$ -2$$for $$/alpha $$ and  $$3$$ for  $$\beta$$ into the general form and use the initial values given in the problem statement to begin to solve for $$ C_1 $$and $$ C_2 $$. 

$$ y(t)=C_1 e ^{-2t} cos(3t) + C_2 e ^{-2t} sin(3t) \ $$  $$ y(0)=C_1 \cancelto{1} {e ^{-2(0)}cos(3(0))} + \cancelto{0} {C_2 e ^{-2(0)} sin(3(0))}= 1$$$$ C_1=1$$ 

$$ y'(t)= (-2)(1) e ^{-2t} cos(3t) -(3)(1)e ^{-2t} sin(3t)-(-2) C_2 e ^{-2t} sin(3t) +(3) C_2 e ^{-2t}cos(3t) \ $$  $$ y'(0)= (-2)(1)\cancelto{1} {e ^{-2(0)} cos(3(0))} - \cancelto{0} {3)(1)e ^{-2(0)} sin(3(0))} -\cancelto{0} {(-2) C_2 e ^{-2(0)} sin(3(0))} +(3) C_2 \cancelto{1} { ^{-2(0)}cos(3(0))}=0$$ $$ y'(0)=-2+3C_2=0 \ $$ <br\>

$$ C_2 = \frac{2} {3} \ $$ <br\><br\> Therefore the the solution particular to the given initial conditions is:<br\> $$ y(t)=e ^{-2t} cos(3t) +\frac{2} {3} e ^{-2t} sin(3t) \ $$ <br\> --Nicolas Ellis 05:13, 7 February 2012 (UTC)

Effort Distribution
--Corey Williams 20:40, 3 February 2012 (UTC)