Answer
$y(x)=c_1e^{-x}+c_2e^{-3x}+c_3e^{2x}+\frac{13}{27}-\frac{10}{9}x-\frac{2}{3}x^2$
Work Step by Step
Given: $y'''+2y''-5y'-6y=4x^2$
Solve the auxiliary equation for the differential equation.
$r^3+2r^2-5r-6=0$
Factor and solve for the roots.
$(r+1)(r+3)(r-2)=0$
Roots are: $r_1=-1$, as a multiplicity of 1, $r_2=-3$ as a multiplicity of 1 and $r_3=2$ as a multiplicity of 1.
The general solution is
$y(x)=c_1e^{-x}+c_2e^{-3x}+c_3e^{2x}$
We have $F(x)=4x^2$
Obtain:
$(D^2+2D^2-5R-6)D^3y_p(x)=4x^2$
Therefore, the general solution for $(D^2+2D^2-5R-6)D^3y_p(x)=0$ is:
$y(x)=c_1e^{-x}+c_2e^{-3x}+A_0+A_1x+A_2x^2$
The trial solution can be computed as by plugging back into the given differential equation.
$y_p(x)=A_0+A_1x+A_2x^2$
So, we have:
$D^3(D+1)(D+3)(D-2)y_p(x)=4x^2\\
D^3(D+1)(D+3)(D-2)(A_0+A_1x+A_2x^2)=4x^2$
On comparing coefficients, we get:
$A_0=\frac{13}{27}\\
A_1=-\frac{10}{9}\\
A_2=-\frac{2}{3}$
Therefore, the general solution for the given differential equation is:
$y(x)=c_1e^{-x}+c_2e^{-3x}+c_3e^{2x}+\frac{13}{27}-\frac{10}{9}x-\frac{2}{3}x^2$
