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.

Printable View

- Jul 6th 2008, 05:48 PMlord12differential 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. - Jul 6th 2008, 06:24 PMparticlejohn
- Jul 6th 2008, 10:53 PMkalagota
first, where does this come from?

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