exact solution to differetial equation

Question

dy/dy+xy=x , y(0)=1/2

Answer 1

y'+xy=x

Use integrating factor e^(int x)= e^(x^2/2)

 

y'e^(x^2/2) +xye^(x^2/2)= xe^(x^2/2)

Integrate to get

ye^(x^2/2)=e^(x^2/2) +C

 

y= 1- Ce^(-x^2/2)

y(0)= 1/2

1/2= 1-C

C=1/2

y= 1-1/2 e^(-x^2/2)

Answer 2

dy/dx+ xy = x

dy/dx = x- xy

          = x(1-y)

 

So its variables are separable 

 dy/(1-y) = xdx

Substituting (1-y) = t

                      -dy =dt

                        dy= -dt

                 dy/(1-y) = -dt/t

 

-dt/t =xdx                 : on integrating

-ln t = x²/2 +c

-ln (1-y) = x²/2 + c

 

 

ln (1-y) = - (x²/2 + c)

 

1-y =e^{-((x^2)/2+c)} 

   y= 1-e^{-((x^2)/2+c)} 

 

y= 1- e^ce^{-(x^2)/2    ;} replacing e^c with other constant C

 

y= 1-Ce^{-(x^2)/2}  

Apply initial condition, find and then replace the value of constant C to get the exact solution.