MAT244-2018S > Final Exam

FE-P2

(1/1)

Victor Ivrii:
Find the general solution by method of the undetermined coefficients:
\begin{equation*}
y'''-2y''+4y'-8y= 16 e^{2t} + 30\cos(t);
\end{equation*}

Tim Mengzhe Geng:
First we find the solution for the homogeneous system

y^{(3)}-2y^{(2)}+4y^{(1)}-8y=0

The corresponding characteristic equation is

r^3-2r^2+4r-8=0

Three roots are

r_1=2

r_2=2i

r_3=-2i

Then the solution for the homogeneous system is

y_c(t)=c_1e^{2t}+c_2\cos(2t)+c_3\sin(2t)

Then we follow to find a particular solution $Y(t)$
We should have

Y(t)=Y_1(t)+Y_2(t)

where

Y_1(t)=Ate^{2t}

and

Y_2(t)=M\sin(t)+N\cos(t)

By plugging in to the equation, we can find that $A=2$, $M=2$ and $N=-4$
In this way we get the required general solution

y(t)=c_1e^{2t}+c_2\cos(2t)+c_3\sin(2t)+2te^{2t}+2\sin(t)-4\cos(t)

Meng Wu:
Most of the answer is correct, there is one small computational error.$\\$
For $Y_2(t)$, it should be $Y_2(t)=-4\cos(t)+2\sin(t)$.$\\$

Tim Mengzhe Geng:

--- Quote from: Meng Wu on April 11, 2018, 11:29:00 PM ---Most of the answer is correct, there is one small computational error.$\\$
For $Y_2(t)$, it should be $Y_2(t)=-4\cos(t)+2\sin(t)$.$\\$

--- End quote ---
Thanks and I will modify it