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

First, we find the characteristic equation, which is given by $r^2+6r+9=0$. This equation factors as $(r+3)^2=0$, so $r=-3$ is a double root.

Therefore, the complementary solution is $y_c(x)=c_1e^{-3x}+c_2xe^{-3x}$.

Next, we need to find the particular solution.

Since the right-hand side of the differential equation is $2+x$, we try a particular solution of the form $y_p(x)=Ax+Bx+C$.

Taking the first and second derivatives of $y_p(x)$, we get:

$$y'_p(x)=A+B$$

$$y''_p(x)=0$$

Substituting these into the differential equation, we get:

$$(A+B) + 6(0) + 9(Ax+Bx+C) = 2+x$$

Simplifying, we get:

$$(9A)x + (9B)x + (9C+A+B) = 2+x$$

We need to choose $A$, $B$, and $C$ such that this equation is true for all values of $x$.

Comparing coefficients, we get the following system of equations:

$$9A+9B=1$$

$$9C+A+B=2/9$$

Solving this system of equations gives $A=1/9$, $B=0$, and $C=4/27$. Therefore, the particular solution is $y_p(x)=\frac{1}{9}x+\frac{4}{27}$.

Finally, the general solution is $y(x)=y_c(x)+y_p(x)=c_1e^{-3x}+c_2xe^{-3x}+\frac{1}{9}x+\frac{4}{27}$, which is given by choice (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 $y ^ { tt} + 6 y ^ { t } + 9 y = 2 + x$