Thread: 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.

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.

.

How did you get ?

According to your substitution, .

first, where does this come from?

or is it included in the given?