Toronto Math Forum
MAT2442018S => MAT244Tests => Term Test 2 => Topic started by: Victor Ivrii on March 21, 2018, 02:56:30 PM

Consider equation
\begin{equation}
y'''3y'+2y= 18e^{2t}.
\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).

Part (a)$\\$
The equation for the Wronskian would be:
$$W = c\times exp[\int p_{1}(t)dt] $$
Since there is no $y''$ term, $p_{1}(t)$ would be $0$:
$$W = c\times exp[\int 0 dt]=c\times exp(0)=c$$
Therefore, the Wronskian would be a constant.
Part (b)$\\$
Consider the homogeneous equation:
$$y'''  3y' + 2y = 0$$
The characteristic equation would be:
$$r^33r+2=0$$
Solving this gives:
$$(r1)^2(r+2)\rightarrow r_{1}=r_{2}=1, r_{3}=2$$
Therefore, the homogeneous solution would be:
$$y_{c}(t)=c_{1}e^t+c_{2}te^t+c_{3}e^{2t}\qquad(2)$$
Computing the Wronskian:
$$W=\begin{array}{c c c}e^t &te^t &e^{2t}\\e^t&(t+1)e^t&2e^{2t}\\e^t&(t+2)e^t&4e^{2t}\end{array}=4(t+1)+2(t+2)t(4+2)+(t+2)(t+1)=9$$
Therefore, the Wronskian is a constant just as expected based on part (a).
Part (c)$\\$
The particular solution should be of the form:
$$Y(t)=Ae^{2t}$$
Since $e^{2t}$ is part of the homogeneous solution, we look for solutions of the form:
$$Y(t)=Ate^{2t}\qquad(3)$$
Differentiating this:
$$Y'(t)=Ae^{2t}2Ate^{2t} \qquad(4)$$
Differentiating again:
$$Y''(t)=4Ae^{2t}+4Ate^{2t}$$
Differentiating once more:
$$Y'''(t)=12Ae^{2t}8Ate^{2t} \qquad(5)$$
Plugging (3), (4) and (5) into (1):
$$12Ae^{2t}8Ate^{2t}3Ae^{2t}+6Ate^{2t}+2Ate^{2t}=18e^{2t}$$
Simplifying gives:
$$9Ae^{2t}=18e^{2t}$$
Therefore, $A=2$. Subbing this value of A into (3) and combining it with (2) gives the general solution:
$$y(t)=c_{1}e^t+c_{2}te^t+c_{3}e^{2t}+2te^{2t}$$