Solve the differential equation using the method of variation of parameters. $$y ^ { tt } - y ^ { t } = e ^ { 9 x }$$
A)$y ( x ) = c _ { 1 } + c _ { 2 } x e ^ { x } + \frac { e ^ { 9 x } } { 72 }$
B) $y ( x ) = c _ { 1 } + x e ^ { 9 x }$
C) $y ( x ) = c _ { 1 } + ( 1 + x ) x e ^ { 9 x }$
D)$y ( x ) = c _ { 1 } + c _ { 2 } e ^ { x } + \frac { e ^ { 9 x } } { 72 }$
E) $y ( x ) = c _ { 1 } + c _ { 2 } x e ^ { 9 x }$

Question

Quizplus · Accepted Answer

The characteristic equation is $r^2-r=0$, so its roots are $r=0$ and $r=1$. 
Thus, the complementary solution is $y_c(x) = c_1 +c_2 e^x$. 
We need to find the particular solution $y_p(x)$ using the method of variation of parameters. We assume that $y_p(x)=u_1(x)y_1(x)+u_2(x)y_2(x)$ where $y_1(x)$ and $y_2(x)$ are solutions of the homogeneous equation and $u_1(x)$ and $u_2(x)$ are functions to be determined. 
Here, $y_1(x)=1$ and $y_2(x)=e^x$. 
We find that $y_1(x)$ and $y_2(x)$ are linearly independent. 
Then, we have:
\begin{align*}
y_p(x) &= u_1(x)y_1(x)+u_2(x)y_2(x)\
&= u_1(x)+u_2(x)e^x
\end{align*}
We find $u_1(x)$ and $u_2(x)$ by substituting $y_p(x)$ and its derivatives into the differential equation and equating the coefficient of $e^x$ and the constant term to zero, respectively. 
We get:
\begin{align*}
u_1'(x)+u_2'(x)e^x&=0\
u_1'(x)+u_2'(x)e^x&=e^{9x}
\end{align*}
Solving for $u_1'(x)$ and $u_2'(x)$, we get:
\begin{align*}
u_1'(x)&=-\frac{1}{e^x}\
u_2'(x)&=\frac{e^{8x}}{e^x}
\end{align*}
Integrating, we get:
\begin{align*}
u_1(x)&=-e^{-x}+C_1\
u_2(x)&=\frac{1}{9}e^{8x}+C_2
\end{align*}
Therefore, the general solution is:
\begin{align*}
y(x)&=y_c(x)+y_p(x)\
&=c_1+c_2e^x+(-e^{-x}+C_1)e^{-x}+\left(\frac{1}{9}e^{8x}+C_2ight)e^x\
&=c_1+C_1e^{-x}+C_2e^{8x}+\frac{1}{9}e^{9x}
\end{align*}
Since $C_1$ and $C_2$ are constants of integration, we can combine them into one constant $c_3$. Thus, the final solution is:
$$y(x)=c_1+c_3e^{-x}+\frac{1}{9}e^{9x}$$
This matches option D.

Solve the Differential Equation Using the Method of Variation of Parameters