Question

Solve y"' -5y"+ 4y'=8ex + 4x.

Answer 1

Let $y'=z.$ Then

$$z''-5z'+4z=8e^x+4x$$

The homogeneous equation is

$$z''-5z'+4z=0$$

The characteristic equation is

$$r^2-5r+4=0$$

$$r_1=1, r_2=4$$

The general solution of the homogeneous differential equation is

$$z_h=C_1e^x+C_2e^{4x}$$

Find the particular solution of the nonhomogeneous equation in form

$$z_p=Axe^x+Bx+C$$

Then

$$z_p'=Ae^x+Axe^x+B$$

$$z_p''=2Ae^x+Axe^x$$

Substitute

$$2Ae^x+Axe^x-5Ae^x-5Axe^x-5B$$

$$+4xAe^x+4Bx+4C=8e^x+4x$$

$$xe^x:0=0$$

$$e^x:-3A=8$$

$$x^1:4B=4$$

$$x^0:-5B+4C=0$$

$$z_p=-\dfrac{8}{3}xe^x+x+\dfrac{5}{4}$$

The general solution of the differential equation

$$z''-5z'+4z=8e^x+4x$$

is

$$z=C_1e^x+C_2e^{4x}-\dfrac{8}{3}xe^x+x+\dfrac{5}{4}$$

Integrate

$$y=\int(C_1e^x+C_2e^{4x}-\dfrac{8}{3}xe^x+x+\dfrac{5}{4})dx$$

$$\int xe^xdx=xe^x-e^x+C_3$$

$$y=c_1+c_2e^x+c_3e^{4x}-\dfrac{8}{3}xe^x+\dfrac{1}{2}x^2+\dfrac{5}{4}x$$