# Thread: Help solving: y" = y' * e^y

1. ## Help solving: y" = y' * e^y

Initial conditions: y(0) = 0, y'(0) = 1

I asked a question this morning which was promptly answered by mr. fantastic (thanks again if he is reading this).

I have come to the end of my problem set and the only problem remaining is the one in the thread title.

I have done similar problems involving setting u = y' and applying the chain rule, and then performing integration by means of separable variables with the initial conditions. Basically, I think I could complete it mostly on my own, but I can't quite figure out what to do with the e^y term. If anyone could provide some insight it would be appreciated. Thanks.

2. Originally Posted by LucasV
Initial conditions: y(0) = 0, y'(0) = 1

I asked a question this morning which was promptly answered by mr. fantastic (thanks again if he is reading this).

I have come to the end of my problem set and the only problem remaining is the one in the thread title.

I have done similar problems involving setting u = y' and applying the chain rule, and then performing integration by means of separable variables with the initial conditions. Basically, I think I could complete it mostly on my own, but I can't quite figure out what to do with the e^y term. If anyone could provide some insight it would be appreciated. Thanks.
Substitute v = dy/dx and note that $\displaystyle \frac{dv}{dx} = \frac{dv}{dy} \cdot \frac{dy}{dx} = \frac{dv}{dy} \cdot v$. Solve for v and use this solution to solve for y.

3. Alright, so I let u = y', which makes y" = u * du/dy....

which changes the eq to:

u * du/dy = u * e^y (u's cancel)

I apply the method of separable variables and this gives:

u = e^y + c

when x=0, y(0)=0, y'(0)=1

therefore 1 = e^0 + c which yields c = 0

I then change u back to dy/dx and the eq becomes:

dy/dx = e^y

by performing separable variables again, this yields

-1/(e^y) = x, solving for y gives -> y = ln(-1/x) + c

this seems a bit off to me because solving for c requires taking the natural log of infinity which doesn't help.

have I made a mistake somewhere?

4. Originally Posted by LucasV
Initial conditions: y(0) = 0, y'(0) = 1

I asked a question this morning which was promptly answered by mr. fantastic (thanks again if he is reading this).

I have come to the end of my problem set and the only problem remaining is the one in the thread title.

I have done similar problems involving setting u = y' and applying the chain rule, and then performing integration by means of separable variables with the initial conditions. Basically, I think I could complete it mostly on my own, but I can't quite figure out what to do with the e^y term. If anyone could provide some insight it would be appreciated. Thanks.
You could simply integrate both sides with respect to $\displaystyle \displaystyle x$...

$\displaystyle \displaystyle \frac{d^2y}{dx^2} = e^y\,\frac{dy}{dx}$

$\displaystyle \displaystyle \int{\frac{d^2y}{dx^2}\,dx} = \int{e^y\,\frac{dy}{dx}\,dx}$

$\displaystyle \displaystyle \int{\frac{d^2y}{dx^2}\,dx} = \int{e^y\,dy}$

$\displaystyle \displaystyle \frac{dy}{dx} = e^y + C_1$

$\displaystyle \displaystyle \frac{dx}{dy} = \frac{1}{e^y + C_1}$

$\displaystyle \displaystyle \frac{dx}{dy} = \frac{e^y}{e^{2y} + C_1e^y}$

$\displaystyle \displaystyle \int{\frac{dx}{dy}\,dy} = \int{\left(\frac{1}{e^{2y} + C_1e^y}\right)e^y\,dy}$

$\displaystyle \displaystyle x = \int{\frac{1}{u^2 + C_1u}\,du}$ after making the substitution $\displaystyle \displaystyle u = e^y$. You can now solve using partial fractions.

5. Thanks to both of you. For some reason I couldn't see that I should have found my c2 PRIOR to taking ln of both sides. If I take my solution now and stuff it into the ode it yields the desired result, with some very light algebraic manipulation. Solved.

6. Originally Posted by LucasV
Alright, so I let u = y', which makes y" = u * du/dy....

which changes the eq to:

u * du/dy = u * e^y (u's cancel)

I apply the method of separable variables and this gives:

u = e^y + c

when x=0, y(0)=0, y'(0)=1

therefore 1 = e^0 + c which yields c = 0

I then change u back to dy/dx and the eq becomes:

dy/dx = e^y

by performing separable variables again, this yields

-1/(e^y) = x, solving for y gives -> y = ln(-1/x) + c

this seems a bit off to me because solving for c requires taking the natural log of infinity which doesn't help.

have I made a mistake somewhere?