## Exercise:

Find the general solution of the linear equation with constant factors

$$y” -5y’+6y=e^{-2x}$$

## Solution:

The homogeneous equation is:

$$y” -5y’+6y=0$$

The characteristic polynomial is:

$$\chi(\lambda)=\lambda^2 -5\lambda +6 = (\lambda-2)(\lambda-3)$$

This gives the two eigenvalues:

$$\lambda_1 = 2 \qquad \lambda_2 = 3$$

This yields the fundamental system:

$$y_1=e^{2x} \qquad y_2=e^{3x}$$

The non-homogeneous equation is:

$$y” -5y’+6y=e^{-2x}$$

The perturbation is $$e^{-2x}$$ and since $$\lambda = -2$$ is not a root of $$\chi(\lambda)$$, there is no resonance.

For the particular solution we have:

$$y_s = A e^{-2x}$$

Pluggin in an comparing the coefficients implies:

$$A=\frac{1}{20}$$

Therefore the general solution is:

$$y=y_s + C_1~y_1+C_2~y_2\\\\ y=\frac{1}{20}~e^{-2x}+C_1~e^{2x}+C_2~e^{3x}$$