Question

Show that the general solution of the differential equation

4z'' - 4z' - 3z = cos2x takes the expression: z = c1e^(3x/2) + c2e^(-x/2) - (19/425)cos2x - (8/425)sin2x

Answer 1

Solution.

$$4z''-4z'-3z=\cos{2x}.$$

This is a linear differential equation of the second-order with constant coefficients. The general solution of this equation is found as the sum of the general solution $\tilde{z}$ ​ of the corresponding homogeneous equation and some particular integral solution $z_*​$ of the inhomogeneous equation:$zy=\tilde{z}+z_*.$

Сompose and solve the characteristic equation:

$$4\lambda^2-4\lambda-3=0, \newline \lambda_1=-\frac{1}{2}, \lambda_2=\frac{3}{2}.$$

 Therefore, $\tilde{z}=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}.$ So,

$$\tilde{z}=C_1e^{-\frac{1}{2}x}+C_2e^{\frac{3}{2}x}.$$

Since the right-hand side contains a $\cos{2x}$ , we find a particular integral solution $z_*$ ​ ​in the form: 

$$z_*=A\cos{2x}+B\sin{2x}.$$

Using the method of undetermined coefficients we find that $A=-\frac{19}{425}, B=-\frac{8}{425}.$

So, the particular integral solution of equation $4z''-4z'-3z=\cos{2x}$ is

$$z_*=-\frac{19}{425}\cos{2x}-\frac{8}{425}\sin{2x}.$$

And the general solution of the same equation is

$$z=C_1e^{-\frac{1}{2}x}+C_2e^{\frac{3}{2}x}-\frac{19}{425}\cos{2x}-\frac{8}{425}\sin{2x}.$$

Answer. $z=C_1e^{-\frac{1}{2}x}+C_2e^{\frac{3}{2}x}-\frac{19}{425}\cos{2x}-\frac{8}{425}\sin{2x}.$