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
#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!