# Math Help - differential equation

1. ## differential equation

solve dy/dx = (y^2)(e^2y), y(0) = 1.

e^2y = x
lnx = 2y.
lnx/2 = y

dy/y^2 = (0.5)(lnx)dx

-y^-1 = (.5)(xlnx-x) + C

C = -1

y = -2/(xlnx -x -1)

Is this correct?

Im stuck here since lnx is not determinable.

2. Originally Posted by lord12
solve dy/dx = (y^2)(e^2y), y(0) = 1.

e^2y = x
lnx = 2y.
lnx/2 = y

dy/y^2 = (0.5)(lnx)dx

-y^-1 = (.5)(xlnx-x) + C

C = -1

y = -2/(xlnx -x -1)

Is this correct?

Im stuck here since lnx is not determinable.
$\frac{dy}{dx} = y^{2}e^{2y}, \ y(0) = 1$

$\ln x = 2y \implies y = \frac{\ln x}{2}$.

How did you get $\frac{dy}{y^2} = \frac{1}{2} \ln x \ dx$?

According to your substitution, $\frac{dy}{y^2} = x \ dx$.

3. first, where does this come from?

$x = e^{2y}$ or is it included in the given?