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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"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran

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?