Question 705055

1. Complementary Function ($y_c$):

The homogeneous equation is $\frac{d^2y}{dx^2} - y = 0$.

Its auxiliary equation is:

$m^2 - 1 = 0 \implies m = \pm 1$

So, the complementary function is:

$y_c = C_1e^x + C_2e^{-x}$

2. Particular Integral ($y_p$):

Let the particular solution be of the form:

$y_p = (Ax + B)\sin x + (Cx + D)\cos x$

Differentiating $y_p$ with respect to $x$:

$y_p' = A\sin x + (Ax+B)\cos x + C\cos x - (Cx+D)\sin x$

$y_p' = (A - Cx - D)\sin x + (C + Ax + B)\cos x$

Differentiating again:

$y_p'' = -C\sin x + (A-Cx-D)\cos x + A\cos x - (C+Ax+B)\sin x$

$y_p'' = (-2C - Ax - B)\sin x + (2A - Cx - D)\cos x$

Substitute $y_p$ and $y_p''$ into the original differential equation ($y_p'' - y_p = x\sin x$):

$[(-2C - Ax - B)\sin x + (2A - Cx - D)\cos x] - [(Ax + B)\sin x + (Cx + D)\cos x] = x\sin x$

$(-2Ax - 2B - 2C)\sin x + (-2Cx + 2A - 2D)\cos x = x\sin x$

Comparing coefficients on both sides:

For $x\sin x$: $-2A = 1 \implies A = -\frac{1}{2}$

For $\sin x$: $-2B - 2C = 0 \implies B = -C$

For $x\cos x$: $-2C = 0 \implies C = 0 \implies B = 0$

For $\cos x$: $2A - 2D = 0 \implies D = A = -\frac{1}{2}$

So, $A = -\frac{1}{2}, B = 0, C = 0, D = -\frac{1}{2}$.

Thus, $y_p = -\frac{1}{2}x\sin x - \frac{1}{2}\cos x$

General Solution: $y = y_c + y_p$

$y = C_1e^x + C_2e^{-x} - \frac{1}{2}(x\sin x + \frac{1}{2}\cos x)$