Solve by using Undetermined Coefficients:

y''+y=e^x+x^3 , y(0) = 2 , y'(0) = 0

so I found r to equal + i which gives the first particular solution
y(x) = C1*cosx+C2sinx

for the first side I got yp(x) = Ae^x and both y' and y'' to equal Ae^x.

I solved for A to get 1/2.

can someone show me the 2nd part with the x^3? what would my yp(x) be? thanks

With the x^3 look for a solution of the form Ax^3+Bx^2+Cx+D.

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So I used Bx^3+Cx^2+Dx+E since I used A for the first part.

y' = 3Bx^2+2Cx+D
y'' = 6Bx + 2C

(6Bx+2C) + (Bx^3+Cx^2+Dx+E) = x^3

isn't it just x^3(B) = x^3 , B = 1?