Solve the differential equation $(D^2 + 9)y = \cos^3x$.

1. Problem Description

Solve the differential equation $(D^2 + 9)y = \cos^3x$.

2. Solution Steps

The given differential equation is $(D^2 + 9)y = \cos^3x$.

First, we find the complementary function by solving the homogeneous equation $(D^2 + 9)y = 0$. The auxiliary equation is $m^2 + 9 = 0$, which gives $m = \pm 3i$. Thus, the complementary function is $y_c = c_1 \cos(3x) + c_2 \sin(3x)$.

Next, we find the particular integral. Since $\cos^3x = \frac{1}{4}(3 \cos x + \cos 3x)$, the equation becomes $(D^2 + 9)y = \frac{1}{4}(3 \cos x + \cos 3x)$.

We can write the particular integral as $y_p = y_{p1} + y_{p2}$, where $(D^2 + 9)y_{p1} = \frac{3}{4} \cos x$ and $(D^2 + 9)y_{p2} = \frac{1}{4} \cos 3x$.

For $y_{p1}$, we assume $y_{p1} = A \cos x$. Then $D^2 y_{p1} = -A \cos x$.

Substituting into $(D^2 + 9)y_{p1} = \frac{3}{4} \cos x$, we get $(-A \cos x + 9A \cos x) = \frac{3}{4} \cos x$, which gives $8A = \frac{3}{4}$, so $A = \frac{3}{32}$.

Thus, $y_{p1} = \frac{3}{32} \cos x$.

For $y_{p2}$, we assume $y_{p2} = Bx \sin 3x$ (since $\cos 3x$ is a part of the complementary function).

Then $Dy_{p2} = B \sin 3x + 3Bx \cos 3x$, and $D^2 y_{p2} = 3B \cos 3x + 3B \cos 3x - 9Bx \sin 3x = 6B \cos 3x - 9Bx \sin 3x$.

Substituting into $(D^2 + 9)y_{p2} = \frac{1}{4} \cos 3x$, we have $(6B \cos 3x - 9Bx \sin 3x + 9Bx \sin 3x) = \frac{1}{4} \cos 3x$, which simplifies to $6B \cos 3x = \frac{1}{4} \cos 3x$.

Thus, $6B = \frac{1}{4}$, so $B = \frac{1}{24}$.

Therefore, $y_{p2} = \frac{1}{24}x \sin 3x$.

The general solution is $y = y_c + y_{p1} + y_{p2} = c_1 \cos 3x + c_2 \sin 3x + \frac{3}{32} \cos x + \frac{1}{24}x \sin 3x$.

3. Final Answer

$y = c_1 \cos 3x + c_2 \sin 3x + \frac{3}{32} \cos x + \frac{1}{24} x \sin 3x$