1st order differential equation

Question

Solve y'+((1-2x)/x)y=e^(2x)

Answer 1

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

 

There is an integrating factor I(x) that can be applied to both sides of this equation and is the following:

I(x) = e^{∫(1-2x)/x dx}

 

Now ∫(1-2x)/x dx = ∫(1/x - 2)dx = ln(x) - 2x + c

 

Then I(x) = e^{ln(x) - 2x + c} = k *e^ln(x) * e^-2x, where k = e^c

              = kxe^-2x

 

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

 

kxe-^2x(y' + (1 - 2x)/x*y) = kxe-^2x * e^2x

xe^-2xy' + (1 -2x)e^-2xy = x

 

(xe^-2xy)' =x

 

Integrate: xe^-2xy = x²/2 + c

 

y = e^2x/x(x²/2 + c)

   = e^2x((c/x + x/2)

Answer 2

Y=.5xe^(2x)+(ce^(2x))/x