Answer
See below
Work Step by Step
Given $y'''+3y''-4y=0$
Solve the auxiliary equation for the differential equation.
$r^3+3r^2-4r=0$
Factor and solve for the roots.
$(r−1)(r+2)^2=0$
Roots are: $r_1=-2$, as a multiplicity of 2 and $r_2=1$ as a multiplicity of 1.
This implies that there are two independent solutions to the differential equation
$y_1(x)=e^x$
$y_2=e^{-2x}$ and
$y_3(x)=xe^{-2x}$
Therefore, the general equation is equal to $y(x)=C_1e^{-2x}+C_2xe^{-2x}+C_3e^x$
