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

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

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

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