### Author Topic: MT Problem 1  (Read 5696 times)

#### Victor Ivrii

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##### MT Problem 1
« on: March 06, 2013, 09:05:02 PM »
Solve the initial value problem
\begin{equation*}
z'' -3z' + 2z = 2e^{3x} , \qquad z(0) = 1 , \qquad z'(0) = 0 .
\end{equation*}

#### Rudolf-Harri Oberg

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##### Re: MT Problem 1
« Reply #1 on: March 06, 2013, 09:57:28 PM »
We use Milman's method to find a particular solution:
$$L\left[A\frac{x^m}{m!}e^{rx}\right]=Ae^{rx}\left(Q(r)\frac{x^m}{m!}+Q'(r)\frac{x^{m-1}}{(m-1)!}+Q''(r)\frac{x^{m-2}}{2!(m-2)!}+... \right)$$
In this case
$Q=r^2-3r+2, Q'=2r-3, Q''=2$. We want to have $r=3$. So we evaluate that $Q(3)=2, Q'(3)=3, Q''(3)=2$.
Also, let $m=0$. Then $L(Ae^{3x})=2Ae^{3x}$. This implies that $A=1$.
In conclusion, solution is $Y_p=e^{3x}$.

For general solution to homogeneous equation, solve $r^2-3r+2=0$ yielding $r_1=2$, $r_2=1$.
So, $Y_{gen.hom}=c_1e^{2x}+c_2e^{x}$

General solution for the whole system is $Y_G=Y_{gen.hom}+Y_p=c_1e^{2x}+c_2e^{x}+e^{3x}$.
Condition $Z(0)=1$ yields that $c_1+c_2+1=1$. Condition $Z'(0)=0$ yields that $2c_1+c_2+3=0$. Solving these gives $c_1=-3, c_2=3$.

In conclusion we get that: $y=-3e^{2x}+3e^x+e^{3x}$.
« Last Edit: March 06, 2013, 10:41:02 PM by Rudolf-Harri Oberg »

#### Jeong Yeon Yook

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##### Re: MT Problem 1
« Reply #2 on: March 06, 2013, 10:14:12 PM »
#1

#### Branden Zipplinger

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##### Re: MT Problem 1
« Reply #3 on: March 06, 2013, 10:27:36 PM »
here's my solution. sorry for the picture again

#### Devangi Vaghela

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##### Re: MT Problem 1
« Reply #4 on: March 06, 2013, 10:30:46 PM »
This is my solution!
« Last Edit: March 06, 2013, 10:34:19 PM by Devangi Vaghela »

#### Devangi Vaghela

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##### Re: MT Problem 1
« Reply #5 on: March 06, 2013, 10:31:52 PM »
The C is suppose to be a 3 in the initial conditions part!***
« Last Edit: March 06, 2013, 10:34:57 PM by Devangi Vaghela »

#### Victor Ivrii

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##### Re: MT Problem 1
« Reply #6 on: March 07, 2013, 03:32:24 AM »
Rudolf-Harri' solution is perfect. No point to post technically inferior solutions (scanned rather than typed) which do not contribute anything new. Since problem was just for grabs I note all other contributions but that's all.

Branden's picture "me and my solution" is a poster boy for two topics "amazing capabilities of smart phones" and "how not to scan"