Toronto Math Forum
MAT2442018F => MAT244Tests => Term Test 2 => Topic started by: Victor Ivrii on November 20, 2018, 05:55:03 AM

Consider equation
\begin{equation}
y'''y'' +4y'4y= 8\cos(2t).
\label{21}
\end{equation}
(a) Write a differential 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 (\ref{21}).

Here is my answer

There are three parts in this question.

second part

3rd part

a)
The coefficient of $y''$ is 1
then Wronskain is $Ce^{t}$
b)
Use homogeneous equation to find fundamental solutions
$y'''y''+4y'4y=0$
Then $r^3r^2+4r4=0$
Then $(r^2+4)(r1)=0$
Then$ r_1=2i, r_2=2i, r_3=1$
Then the solution is $y=C_1\cos2t+C_2\sin2t+C_3e^t$
So
$W(y_1, y_2, y_3)(t) = \begin{bmatrix} \cos2t&\sin2t&e^{t}\\2\sin2t&2\cos2t&e^{t}\\4\cos2t&4\sin2t&e^{t}\\ \end{bmatrix}=10e^{t}$
This is consistent with what we get in part (a)
c) Use undetermined coefficients method
Assume $y(t) = At\cos{2t}+Bt\sin{2t}$
$y'(t) = A\cos{2t}2At\sin2t+B\sin2t+2Bt\cos2t$
$y''(t) = 4A\sin{2t}4At\cos2t+8At\sin2t8B\sin2t$
$y'''(t) = 12A\cos{2t}+8At\sin2t12B\sin2t8Bt\cos2t$
Plug into the equation,
We get $8A4B=8, 4A8B=0$
we get $A=\frac{4}{5}$
$B=\frac{2}{5}$
Thus, $Y=C_1\cos2t+C_2\sin2t+C_3e^t\frac{4}{5}t\cos2t\frac{2}{5}t\sin2t$