Question

Find the general solution to y"+4y'+3y=x

Answer 1

Characteristic equation: $r^2+4r+3=0\to r_1=-3,r_2=-1.$

General solution of the homogeneous equation $y''+4y'+3y=0$ is $y=C_1e^{-4x}+C_2e^{-x}$ .

The particular solution might have the form: $y_p=Ax+b$.

After substitution $y_p$ into the equation we have: $4A+3Ax+3B=x$.

So, $A=\frac{1}{3},B=-\frac{4}{9}.$

General solution of our ODE: $y=C_1e^{-4x}+C_2e^{-x}+\frac{x}{3}-\frac{4}{9}.$