Toronto Math Forum
MAT2442018S => MAT244Tests => Final Exam => Topic started 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*}

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^36r^2+11r6=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}

Small Error: $W_2(x)$ should be $2e^{4x}$.

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 $(1)^{i+j}$

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.

Also for $(20)$, $\ln(e^x)$ can be simplified as $\ln(e^x)=x$.

Also for $(20)$, $\ln(e^x)$ can be simplified as $\ln(e^x)=x$.
Thanks again and I will modify it :)

Since the solution is incomplete after Y(x),
I am attaching a copy of my solution

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)

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 $c_1$, $c_2$,...