Answer to Question #239722 in Differential Equations for sam

Question

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

Answer 1

Genral solution $y(x)$ is the sum of the general solution $y_g(x)$ of homogeneous equation $y'' + 4y' + 3y =0$ and the partial solution $y_p(x)$ of non-homogeneous equation $y'' + 4y' + 3y = x$:

$$y(x) = y_g(x) + y_p(x)$$

Let the partial solution be $y_p(x) = \dfrac13x-\dfrac49$ (one can verify by direct substitution).

To find the general solution let's solve the characteristic equation:

$$\lambda^2 +4\lambda+3=0\\ \lambda_1 = -1,\space \lambda_2 = -3$$

Thus, obtain:

$$y_g(x) = Ae^{\lambda_1x} + Be^{\lambda_2x} = Ae^{-x} + Be^{-3x}$$

Finally:

$$y(x) = Ae^{-x} + Be^{-3x} + \dfrac13x-\dfrac49$$

Answer. $y(x) = Ae^{-x} + Be^{-3x} + \dfrac13x-\dfrac49$.