Solving this linear D.E. using an integrating factor?

Question

Ok....so D.E. is: #y'+3x^(2)y=x^2#

I know that the integrating factor is #3x^2#

I then get to this point...

#e^(x^3)y= ∫x^(2)e^(x^(3))#

I thought to do a u-sub, having

#u=x^(3)# and #(1/3)du= x^(2)#

However, the answer is:

#y=(1/3)+ce^(-x^3)#

And the continuation of the u-sub results in #e^(x^3)y=(1/3)e^(x^3)+c#

Solving for y does not satisfy. The u-sub seemed to work out, so I didn't think that I needed to use integration by parts...but if that is the case, I could use a nudge in the right direction.

Thank you.

901 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-05T10:37:13

#y(x) = 1/3 +Ce^(-x^3)#

The equation is in the form:

#y'(x) + a(x) y(x) = f(x)#

Find a primitive of #a(x)#:

#A(x) = int_0^x a(t)dt = int_0^x 3t^2dt = x^3#

The integrating factor is:

#I(x) = e^(x^3)#

Multiply both sides of the equation by #I(x)#:

#e^(x^3)y'(x) + 3x^2e^(x^3)y(x) = x^2e^(x^3)#

Now the first member is:

#d/dx (e^(x^3)y(x)) = x^2e^(x^3)#

So:

#e^(x^3)y(x) = int x^2e^(x^3) dx = 1/3 int e^(x^3) d(x^3) =e^(x^3)/3+C#

#y(x) = 1/3 +Ce^(-x^3)#

and in fact:

#y'(x) +3x^2y(x) = -3Cx^2e^(-x^3) + x^2 +3Cx^2 e^(-x^3) = x^2#