Solve the differential equation using the method of undetermined coefficients.
$$y ^ { tt } + 6 y ^ { t } + 9 y = 2 + x$$
A)
$y ( x ) = c _ { 1 } e ^ { - 3 x } + c _ { 2 } x e ^ { - 3 x } + \frac { 2 } { 9 } x$
B) $y ( x ) = c _ { 1 } e ^ { - 3 x } + c _ { 2 } x e ^ { - 3 x }$
C)
$y ( x ) = c _ { 1 } e ^ { - 3 x } + \frac { 1 } { 9 } x + \frac { 4 } { 27 }$
D)
$y ( x ) = c _ { 1 } x e ^ { - 3 x } + c _ { 2 } x ^ { 2 } e ^ { - 3 x } + \frac { 4 } { 27 } x$
E)
$y ( x ) = c _ { 1 } e ^ { - 3 x } + c _ { 2 } x e ^ { - 3 x } + \frac { 1 } { 9 } x + \frac { 4 } { 27 }$

Question

Quizplus · Accepted Answer

The characteristic equation is $r^2+6r+9=(r+3)^2=0$. So, the complementary solution is $y_c=c_1e^{-3x}+c_2xe^{-3x}$.
The particular solution is of the form $y_p =Ax+B$, where $A$ and $B$ are constants to be determined. Then, we have $y_p'=A$ and $y_p''=0$. Substituting into the ODE, we get
$$0+6A+9(Ax+B)=2+x.$$
Matching coefficients, we have $9A=0$ and $6A+9B=2$. Solving for $A$ and $B$, we get $A=0$ and $B=\frac{2}{9}$. Therefore, the general solution is
$$y(x)=y_c+y_p=c_1e^{-3x}+c_2xe^{-3x}+\frac19(x+4).$$
Using the initial conditions (not given in the problem), we can find the specific values of $c_1$ and $c_2$. However, since the problem only asks for the best choice, the answer is E.

Solve the Differential Equation Using the Method of Undetermined Coefficients ytt+6yt+9y=2+xy ^ { tt } + 6 y ^ { t } + 9 y = 2 + xytt+6yt+9y=2+x

Solve the Differential Equation Using the Method of Undetermined Coefficients $y ^ { tt } + 6 y ^ { t } + 9 y = 2 + x$