The problem is to solve the second-order non-homogeneous linear differential equation: $y'' - 3y' = x^3$

1. Problem Description

The problem is to solve the second-order non-homogeneous linear differential equation:

$y'' - 3y' = x^3$

2. Solution Steps

First, we find the homogeneous solution.

The characteristic equation is $r^2 - 3r = 0$, which factors as $r(r-3) = 0$.

The roots are $r_1 = 0$ and $r_2 = 3$.

The homogeneous solution is $y_h = c_1 e^{0x} + c_2 e^{3x} = c_1 + c_2 e^{3x}$.

Next, we find a particular solution.

Since the right-hand side is $x^3$, we would normally try a polynomial of the form $Ax^3 + Bx^2 + Cx + D$. However, since $r=0$ is a root of the characteristic equation, we must multiply by $x$. So, we try a particular solution of the form:

$y_p = x(Ax^3 + Bx^2 + Cx + D) = Ax^4 + Bx^3 + Cx^2 + Dx$

$y_p' = 4Ax^3 + 3Bx^2 + 2Cx + D$

$y_p'' = 12Ax^2 + 6Bx + 2C$

Substituting into the differential equation:

$(12Ax^2 + 6Bx + 2C) - 3(4Ax^3 + 3Bx^2 + 2Cx + D) = x^3$

$12Ax^2 + 6Bx + 2C - 12Ax^3 - 9Bx^2 - 6Cx - 3D = x^3$

$-12Ax^3 + (12A - 9B)x^2 + (6B - 6C)x + (2C - 3D) = x^3$

Comparing coefficients:

$-12A = 1$, so $A = -\frac{1}{12}$

$12A - 9B = 0$, so $12(-\frac{1}{12}) - 9B = 0$, which gives $-1 - 9B = 0$, so $B = -\frac{1}{9}$

$6B - 6C = 0$, so $6(-\frac{1}{9}) - 6C = 0$, which gives $-\frac{2}{3} - 6C = 0$, so $C = -\frac{1}{9}$

$2C - 3D = 0$, so $2(-\frac{1}{9}) - 3D = 0$, which gives $-\frac{2}{9} - 3D = 0$, so $D = -\frac{2}{27}$

Thus, $y_p = -\frac{1}{12}x^4 - \frac{1}{9}x^3 - \frac{1}{9}x^2 - \frac{2}{27}x$

The general solution is $y = y_h + y_p = c_1 + c_2 e^{3x} - \frac{1}{12}x^4 - \frac{1}{9}x^3 - \frac{1}{9}x^2 - \frac{2}{27}x$

3. Final Answer

$y = c_1 + c_2 e^{3x} - \frac{1}{12}x^4 - \frac{1}{9}x^3 - \frac{1}{9}x^2 - \frac{2}{27}x$