MAT244-2018S > Final Exam

FE-P2

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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
\begin{equation}
    y^{(3)}-2y^{(2)}+4y^{(1)}-8y=0
\end{equation}
The corresponding characteristic equation is
\begin{equation}
    r^3-2r^2+4r-8=0
\end{equation}
Three roots are
\begin{equation}
    r_1=2
\end{equation}
\begin{equation}
    r_2=2i
\end{equation}
\begin{equation}
    r_3=-2i
\end{equation}
Then the solution for the homogeneous system is

\begin{equation}
    y_c(t)=c_1e^{2t}+c_2\cos(2t)+c_3\sin(2t)
\end{equation}
Then we follow to find a particular solution $Y(t)$
We should have
\begin{equation}
    Y(t)=Y_1(t)+Y_2(t)
\end{equation}
where
\begin{equation}
    Y_1(t)=Ate^{2t}
\end{equation}
and
\begin{equation}
    Y_2(t)=M\sin(t)+N\cos(t)
\end{equation}
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
\begin{equation}
    y(t)=c_1e^{2t}+c_2\cos(2t)+c_3\sin(2t)+2te^{2t}+2\sin(t)-4\cos(t)
\end{equation}

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

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