The answer is yt c1exx22xc2 but I am not sure how they got t

The answer is y(t)= c1+e^(x)((x^2)/2)-x+c2) but I am not sure how they got there. Thanks in advance!

Write the general solution to the given ODE. y\'\'-y\'=xe^x The answer is y(t)= c1+e^(x)((x^2)/2)-x+c2) but I am not sure how they got there.

Solution

the general solution= particular solution +complementary solution

for complementary solution:

y \'\' - y \' =0

y = A+B e^x

for particular solution

consider

y \' \' -y \' =xe^x

(D²-D)y=xe^x

y= xe^x /(D²-D) = (1/D-1 - 1/D)xe^x = e^(x)((x^2)/2)-x+c)

hence the general solution is:

y(t)= c1+e^(x)((x^2)/2)-x+c2)   c1,c2 are constants