Yy'-2xe^x =0 using Separation of Variables?

1 Answer
Apr 25, 2018

We have:

#y(dy/dx) = 2xe^x#

#ydy = 2xe^xdx#

The left hand side is easily integrated, however we must use IDP for the right hand side.

Let #u = 2x# and #dv = e^x#.

#int u dv = uv - int vdu#

#int 2xe^x = 2xe^x - int 2e^xdx#

#int 2xe^x = 2xe^x- 2e^x + C#

We can now put this all together.

#1/2y^2 = 2xe^x - 2e^x + C#

#y =+- sqrt(4xe^x - 4e^x + 2C)#

Hopefully this helps!