Toronto Math Forum

MAT244-2018S => MAT244--Tests => Final Exam => Topic started by: Victor Ivrii on April 11, 2018, 08:42:26 PM

Title: FE-P3
Post by: Victor Ivrii on April 11, 2018, 08:42:26 PM
Find the general solution of
\begin{equation*}
y''' -6y'' +11y'- 6y=2\frac{e^{3x}}{e^x+1} .
\end{equation*}
Title: Re: FE-P3
Post by: Tim Mengzhe Geng on April 11, 2018, 11:37:34 PM
First we find the solution for the homogeneous system
\begin{equation}
y^{(3)}-6y^{(2)}+11y^{(1)}-6y=0
\end{equation}
The corresponding characteristic equation is
\begin{equation}
r^3-6r^2+11r-6=0
\end{equation}
Three roots are
\begin{equation}
r_1=1
\end{equation}
\begin{equation}
r_2=2
\end{equation}
\begin{equation}
r_3=3
\end{equation}
Then the solution for the homogeneous system is

\begin{equation}
y_c(t)=c_1e^{x}+c_2e^{2x}+c_3e^{3x}
\end{equation}
where
\begin{equation}
y_1(t)=e^{x}
\end{equation}
\begin{equation}
y_2(t)=e^{2x}
\end{equation}
\begin{equation}
y_3(t)=e^{3x}
\end{equation}
Then we follow to find the required solution to the nonhomogeneous equation. We use Variation of Parameters. We have
\begin{equation}
W[y_1,y_2,y_3]=2e^{6x}
\end{equation}
\begin{equation}
W_1[y_1,y_2y_3]=e^{5x}
\end{equation}
\begin{equation}
W_2[y_1,y_2y_3]=-2e^{4x}
\end{equation}
\begin{equation}
W_3[y_1,y_2，y_3]=e^{3x}
\end{equation}
and
\begin{equation}
g(x)=2\frac{e^{3x}}{e^{x}+1}
\end{equation}
And then we have the following integration
\begin{equation}
\int \frac{W_1\cdot g(x)dx}{W[y_1,y_2,y_3]}=\int\frac{e^{2x}}{e^{x}+1}
\end{equation}
\begin{equation}
\int\frac{e^{2x}}{e^{x}+1}=e^{x}-\ln(e^{x}+1)+c_4
\end{equation}
\begin{equation}
\int \frac{W_2\cdot g(x)dx}{W[y_1,y_2,y_3]}=-2\int\frac{e^{x}}{e^{x}+1}
\end{equation}
\begin{equation}
-2\int\frac{e^{x}}{e^{x}+1}=-2\ln(e^{x}+1)+c_5
\end{equation}
\begin{equation}
\int \frac{W_3 \cdot g(x)dx}{W[y_1,y_2,y_3]}=\int\frac{1}{e^{x}+1}
\end{equation}
\begin{equation}
\int\frac{1}{e^{x}+1}=x-\ln(e^{x}+1)+c_6
\end{equation}
And finally, the required general solution $y(t)$
\begin{equation}
y(t)=\sum_{i=1}^3 y_i(t)\cdot\int\frac {W_i\cdot g(x)dx}{W[y_1,y_2,y_3]}
\end{equation}
Title: Re: FE-P3
Post by: Meng Wu on April 11, 2018, 11:47:31 PM
Small Error: $W_2(x)$ should be $2e^{4x}$.
Title: Re: FE-P3
Post by: Tim Mengzhe Geng on April 11, 2018, 11:49:31 PM
Small Error: $W_2(x)$ should be $2e^{4x}$.
For this case I don't think so since when we expand the matrix, we have to times $(-1)^{i+j}$
Title: Re: FE-P3
Post by: Meng Wu on April 11, 2018, 11:54:19 PM
Small Error: $W_2(x)$ should be $2e^{4x}$.
For this case I don't think so since when we expand the matrix, we have to times $(-1)^{i+j}$

Oh, you're right. My mistake.
Title: Re: FE-P3
Post by: Meng Wu on April 11, 2018, 11:56:10 PM
Also for $(20)$, $\ln(e^x)$ can be simplified as $\ln(e^x)=x$.
Title: Re: FE-P3
Post by: Tim Mengzhe Geng on April 11, 2018, 11:57:57 PM
Also for $(20)$, $\ln(e^x)$ can be simplified as $\ln(e^x)=x$.
Thanks again and I will modify it :)
Title: Re: FE-P3
Post by: Syed Hasnain on April 12, 2018, 04:33:47 PM
Since the solution is incomplete after Y(x),
I am attaching a copy of my solution
Title: Re: FE-P3
Post by: Tim Mengzhe Geng on April 12, 2018, 10:01:46 PM
Since the solution is incomplete after Y(x),
I am attaching a copy of my solution
Sorry what do you mean by "the solution is incomplete after Y(x)"
I did write a bit more on the exam (expanding the summation) but I think one should be able to get full marks if he integrates everything and mention how the solution is composed.(Given that the integral is correct)
Title: Re: FE-P3
Post by: Victor Ivrii on April 15, 2018, 03:15:29 AM
Since the solution is incomplete after Y(x),
I am attaching a copy of my solution
The only thing which was missing in the solution, is the final answer, but it warrants neither such claim, nor uploading your solution.

General remark:
It would be better to denote "parameters" by uppercase letters $C_1(x)$, $C_2(x)$,... and constants by lowercase letters $c_1$, $c_2$,...