Find the general solution?

Question

Find the general solution of y'+2xy=xe^(-x^2).

 

Answer: y=e^(-x^2)(x^2/2+C)

 

Please show all your work step by step.

Answer 1

Sun,

 

Remember. The general solution to a first order linear differential equation is given by

 

y = ( ∫u(x)q(x) dx + c )/u(x)

 

where

 

u(x) = e^{∫p(x) dx}

 

For your differential equation,

 

p(x) = 2x

 

and

 

q(x) = xe^{-x^2}

 

So, our integrating factor u(x) is given as

 

u(x) = e^{∫2x dx}

 

       = e^{x^2}

 

And our solution is given as

 

y = ( ∫e^{x^2}*x*e^{-x^2} dx + C )/(e^{x^2})

 

   = (∫x dx +C)*e^{-x^2}

 

   = (x²/2 + C)*e^{-x^2}