Author Topic: TT2--P2D  (Read 2786 times)

Victor Ivrii

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TT2--P2D
« on: March 21, 2018, 03:00:25 PM »
Consider equation
\begin{equation}
y'''+y''+y'+y= 4e^{-t}.
\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--P2D
« Reply #1 on: March 23, 2018, 07:55:28 PM »
I have attached my solution...
thanks

Victor Ivrii

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Re: TT2--P2D
« Reply #2 on: March 24, 2018, 10:51:13 AM »
OK. I post typed solution

a. $\frac{dW}{W}=-\frac{dt}{W}\implies W=Ce^{-t}$.

b.  Characteristic equation $L(k):=k^3+k^2+k+1= (k+1)(k^2+1)=0\implies k_{1,2}=\pm i$, $k_3=-1$. Then
$y_1=\cos(t)$, $y_2=\sin(t)$, $y_3=e^{-t}$ and
\begin{align*}
W(y_1,y_2, y_3)=&\left|\begin{matrix}
\cos(t) &\sin(t) &e^{-t}\\
-\sin(t) &\cos(t) &-e^{-t}\\
-\cos(t) &-\sin(t) &e^{-2t}
\end{matrix}\right|\overset{(A)}{=}
\left|\begin{matrix}
\cos(t) &\sin(t) &1\\
-\sin(t) &\cos(t) &-1\\
-\cos(t) &-\sin(t) &1
\end{matrix}\right|e^{-t}\\
&\overset{(B)}{=}
\left|\begin{matrix}
\cos(t) &\sin(t) &1\\
-\sin(t) &\cos(t) &-1\\
0 &0 &2
\end{matrix}\right|e^{-t}=2e^{-t}.
\end{align*}

c. General solution of homogeneous equation is $$y^*:= C_1\cos(t) +C_2\sin(t) +C_3e^{-t}.$$ Special solution of is $\bar{y}:= Ate^{-t}$ where $AL'(-1)=4$, $L'(k)=3k^2+2k+1\implies L'(-1)=2\implies A=2$. So
$$
y= 2te^{-t}+ C_1\cos(t)+C_2\sin(t)+C_3e^{-t}.
$$