MAT244-2018S > Term Test 2

TT2--P2M

(1/1)

Victor Ivrii:
Consider equation

y'''-7y'+6y= 100e^{-3t}.
\tag{1}

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:
I have attached my solution ....
Thanks

Victor Ivrii:
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
$$y= 5te^{-3t}+ C_1e^t +C_2e^{2t} +C_3e^{-3t}.$$