Author Topic: TT2--P2M  (Read 2136 times)

Victor Ivrii

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TT2--P2M
« on: March 21, 2018, 03:04:24 PM »
Consider equation
\begin{equation}
y'''-7y'+6y= 100e^{-3t}.
\tag{1}
\end{equation}
a. Write equation for Wronskian of $y_1,y_2,y_3$, which are solutions for homogeneous equation and solve it.

b. Find fundamental system $\{y_1,y_2,y_3\}$ of solutions for homogeneous equation, and find their Wronskian. Compare with (a).

c. Find the general solution of (1).

Syed Hasnain

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Re: TT2--P2M
« Reply #1 on: March 23, 2018, 06:36:32 PM »
I have attached my solution ....
Thanks

Victor Ivrii

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Re: TT2--P2M
« Reply #2 on: March 24, 2018, 11:47:19 AM »
One should not solve the same problem for all sittings (one-trick pony?)

The same solution

a. $\frac{dW}{W}=\frac{dt}{0}\implies W=C$.


b. Characteristic equation $L(k):= k^3-7k+6= (k-1)(k-2)(k+3)=0\implies k_1=1$, $k_2=2$, $k_3=-3$. Then
$y_1=e^t$, $y_2=e^{2t}$, $y_3=e^{-3t}$ and
$$
W(y_1,y_2, y_3)=\left|\begin{matrix}
e^t &e^{2t} &e^{-3t}\\
e^t &2e^{2t} &-3e^{-t}\\
e^t &4e^{2t} &9e^{-2t}
\end{matrix}\right|\overset{(A)}{=}
\left|\begin{matrix}
1 &1 &1\\
1 &2 &-3\\
1 &4 &9
\end{matrix}\right|
\overset{(B)}{=}
\left|\begin{matrix}
1 &1 &1\\
0 &1 &-4\\
0 &3 &8
\end{matrix}\right|=20.
$$

c. General solution of homogeneous equation is $y^*:= C_1e^t +C_2e^{2t} +C_3e^{-3t}$. Special solution  is
$\bar{y}:=Ate^{-3t}$ where $AL'(-3)=4$, $L'(k)=3k^2-7\implies L'(-3)=20\implies A=5$. So
$$
y= 5te^{-3t}+ C_1e^t +C_2e^{2t} +C_3e^{-3t}.
$$

« Last Edit: March 24, 2018, 11:49:52 AM by Victor Ivrii »