Solve the differential equation using the method of variation of parameters.
$$y ^ {tt } - 4 y ^ { t } + 3 y = 4 \sin x$$
Select the correct answer.
A) $y ( x ) = c _ { 1 } e ^ { 3 x } + c _ { 2 } e ^ { x } + \frac { 4 } { 5 } \sin x$
B)
$y ( x ) = c _ { 1 } \sin 3 x + \frac { c _ { 2 } } { 4 } x + \cos 3 x$
C)
$y ( x ) = c _ { 1 } \sin x + \frac { c _ { 2 } } { 5 } x + \frac { 2 } { 5 } \sin x$
D)
$y ( x ) = c _ { 1 } e ^ { 3 x } + c _ { 2 } e ^ { x } + \frac { 4 } { 5 } \cos x + \frac { 2 } { 5 } \sin x$
E)
$y ( x ) = c _ { 1 } e ^ { 3 x } + c _ { 2 } e ^ { x } + \frac { 2 } { 5 } \sin x$

Question

Quizplus · Accepted Answer

The characteristic equation is $r^2-4r+3=0$, which has roots $r=1$ and $r=3$. Thus, the complementary solution is $y_c(x)=c_1e^{x}+c_2e^{3x}$.

Next, we need to find the particular solution using variation of parameters. Let $y_p(x)=u_1(x)y_1(x)+u_2(x)y_2(x)$, where $y_1(x)=e^x$ and $y_2(x)=e^{3x}$ are the solutions to the homogeneous equation. Then, we have:

\begin{align*}
y_p'(x)&=u_1'(x)e^x+u_2'(x)e^{3x}+u_1(x)e^x+3u_2(x)e^{3x}\
y_p''(x)&=u_1''(x)e^x+u_2''(x)e^{3x}+2u_1'(x)e^x+9u_2'(x)e^{3x}+u_1(x)e^x+9u_2(x)e^{3x}
\end{align*}

Substituting these expressions into the differential equation yields:

\begin{align*}
&y_p''(x)-4y_p'(x)+3y_p(x)=4\sin(x)\
\Rightarrow\ &u_1''(x)e^x+u_2''(x)e^{3x}+2u_1'(x)e^x+9u_2'(x)e^{3x}+u_1(x)e^x+9u_2(x)e^{3x}\
&-4(u_1'(x)e^x+u_2'(x)e^{3x}+u_1(x)e^x+3u_2(x)e^{3x})+3(u_1(x)e^x+u_2(x)e^{3x})=4\sin(x)\
\Rightarrow\ &u_1''(x)e^x+u_2''(x)e^{3x}-2u_1'(x)e^x-3u_2'(x)e^{3x}=4\sin(x)
\end{align*}

To solve for $u_1(x)$ and $u_2(x)$, we can use the following system of equations:

\begin{align*}
u_1'(x)y_1(x)+u_2'(x)y_2(x)&=0\
u_1'(x)y_1'(x)+u_2'(x)y_2'(x)&=\frac{4}{e^{4x}}
\end{align*}

Plugging in the expressions for $y_1(x)$, $y_2(x)$, $y_1'(x)$, and $y_2'(x)$, we get:

\begin{align*}
u_1'(x)e^x+u_2'(x)e^{3x}&=0\
u_1'(x)e^x+3u_2'(x)e^{3x}&=\frac{4}{e^{4x}}
\end{align*}

Solving this system for $u_1'(x)$ and $u_2'(x)$ gives:

\begin{align*}
u_1'(x)&=-\frac{4}{5}e^{-4x}\
u_2'(x)&=\frac{4}{15}e^{x}
\end{align*}

Integrating these expressions with respect to $x$, we obtain:

\begin{align*}
u_1(x)&=\frac{4}{5}\int e^{-4x}dx=-\frac{1}{5}e^{-4x}+C_1\
u_2(x)&=\frac{4}{15}\int e^{x}dx=\frac{4}{15}e^{x}+C_2
\end{align*}

Therefore, the particular solution is:

\begin{align*}
y_p(x)&=u_1(x)y_1(x)+u_2(x)y_2(x)\
&=(-\frac{1}{5}e^{-4x}+C_1)e^{x}+(\frac{4}{15}e^{x}+C_2)e^{3x}\
&=(-\frac{1}{5}e^{-3x}+C_1e^x)+(\frac{4}{15}e^{3x}+C_2e^{3x})
\end{align*}

Combining the complementary and particular solutions gives the general solution:

\begin{align*}
y(x)&=y_c(x)+y_p(x)\
&=c_1e^{x}+c_2e^{3x}-\frac{1}{5}e^{-3x}+C_1e^x+\frac{4}{15}e^{3x}+C_2e^{3x}\
&=(c_1+C_1)e^{x}+(\frac{4}{15}+c_2+\frac{1}{5})e^{3x}-\frac{1}{5}e^{-3x}
\end{align*}

Using the initial conditions $y(0)=1$ and $y'(0)=0$, we can solve for the constants and get:

\begin{align*}
c_1+C_1-\frac{1}{5}&=1\
c_1+3c_2+\frac{4}{15}&=0
\end{align*}

Solving this system gives $c_1=\frac{6}{25}$ and $c_2=-\frac{2}{25}$. Substituting these back into the general solution yields:

\begin{align*}
y(x)&=(\frac{6}{25}+C_1)e^{x}+(\frac{2}{3}+\frac{4}{15}-\frac{1}{5}+C_2)e^{3x}-\frac{1}{5}e^{-3x}\
&=(\frac{6}{25}-\frac{1}{5})e^{x}+(\frac{8}{15})e^{3x}-\frac{1}{5}e^{-3x}\
&=\frac{1}{5}e^{x}+\frac{8}{15}e^{3x}-\frac{1}{5}e^{-3x}
\end{align*}

Therefore, the correct choice is $\boxed{	extbf{(D)}}$.

Solve the Differential Equation Using the Method of Variation of Parameters