Answer to Question #346825 in Differential Equations for Genius

Question

Find a general solution for the differential equation -y''=6x+xe^x (a) using the method of undetermined coefficients. (b) using variation of parameters

Answer 1

a)

Homogeneous differential equation is

$$y''=0$$

Characteristic (auxiliary) equation

$$r^2=0$$

The general solution of the homogeneous differential equation is

$$y_h=C_1+C_2x$$

Find the particular solution of the non homogeneous differential equation in the form

$$y_p=x^2(Ax+B)+(Cx+D)e^x$$

Then

$$y_p'=3Ax^2+2Bx+Ce^x+(Cx+D)e^x$$

$$y_p''=6Ax+2B+2Ce^x+Cxe^x+De^x$$

Substitute

$$6Ax+2B+2Ce^x+Cxe^x+De^x=-6x-xe^x$$

$$A=-1, B=0, C=-1, D=2$$

$$C=-1$$

The general solution of the given nonhomogeneous differential equation is

$$y=-x^3-xe^x+2e^x+C_1+C_2x$$

b)

Homogeneous differential equation is

$$y''=0$$

Characteristic (auxiliary) equation

$$r^2=0$$

The general solution of the homogeneous differential equation is

$$y_h=C_1+C_2x$$

$$y'=C_1'+C_2'x+C_2$$

Let

$$C_1'+C_2'x=0$$

Then

$$y'=C_2$$

$$y''=C_2'$$

Substitute

$$C_2'=-6x-xe^x$$

$$C_2=\int(-6x-xe^x)dx$$

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

$$C_2=-3x^2-xe^x+e^x+C_3$$

$$C_1'=-C_2'x$$

$$C_1'=6x^2+x^2e^x$$

$$C_1=\int(6x^2+x^2e^x)dx$$

$$\int x^2e^xdx=x^2e^x-2\int xe^xdx$$

$$=x^2e^x-2xe^x+2e^x+C_4$$

$$C_1=2x^3+x^2e^x-2xe^x+2e^x+C_4$$

The general solution of the given nonhomogeneous differential equation is

$$y=2x^3+x^2e^x-2xe^x+2e^x+C_4$$

$$-3x^3-x^2e^x+xe^x+C_3x$$

$$y=-x^3-xe^x+2e^x+C_4+C_3x$$