# How to solve this non-linear differential equation

I want to solve the following differential equation:

$$y''=e^{x}(y')^2$$

then I substitute $y''=u'u$,$y'=u$ so I got:

$u'=e^{x}u$

But then I don't know how to solve this, may be separate variables or what can be done, Can you help me to solve this please? (In fact I don't know have to solve non-linear differential equations but the substitution was like a hint, so If can you help to clarify the substitution I appreciate it :) )

• well I don't like this, because I have $u=e^{e^{x}}$, and that is not my answer :( – user162343 Jun 6 '15 at 15:30
• You did a mistake in the initial reduction: it should be $u' = e^x u^2 \implies \int \frac{u'}{u^2} dx = \int e^x dx$ – Winther Jun 6 '15 at 15:43
• Ok but it has the same problem right?, because I have to use the exponential – user162343 Jun 6 '15 at 15:45

you can integrate $$\frac{y''}{y'^2} = e^x .$$ it gives $$-\frac 1 {y'} = e^x - c \to y' = \frac{1}{c - e^x}$$ integration the last one gives you $$y = \int \frac{dx}{c-e^x}$$ the last integral can be done with a substitution $$u = c-e^x, du = -e^xdx, dx =\frac{du}{u-c}, \frac{c}{u(u-c)}=\frac1{u-c}-\frac 1u$$ therefore $$y = \int \frac{dx}{c-e^x} = \frac1c\int\left(\frac1{u-c}-\frac 1u\right)=\frac1c\ln\left(\frac{u-c}u\right)=\frac1c\ln\left(\frac{e^x}{e^x-c}\right) + d$$

you need to consider the easier case $c = 0$ separately.

• Ok, so I can rewrite this as $ln(1+ce^{-x})+d$ right? – user162343 Jun 6 '15 at 16:18
• @user162343, yes. you have to be careful now the two $c$'s, one outside and the one inside, have to go together. – abel Jun 6 '15 at 16:20
• Oh right a typo $\frac{1}{c}ln(1+ce^{-x})+d$ – user162343 Jun 6 '15 at 16:21
• @user162343, that is correct. – abel Jun 6 '15 at 16:22
• @user162343, you are welcome. – abel Jun 6 '15 at 16:24

Starting off with a substitution is a good idea.

Now you can use separation of variables. You equation is

$$\frac{du}{dx} = e^xu^2,$$

which can be separated as

$$du/u^2 = e^x dx.$$

Integrate both sides:

$$\int \frac{1}{u^2}du = \int e^x dx.$$

Note that the left hand side is $-1/u$ (the derivative of this function is $1/u^2$). And the right hand side is $e^x+C$ for some arbitrary constant $C$. Thus, $$-\frac{1}{u} = e^x+C$$.

• well I don't like this, because I have $u=e^{e^{x}}$, and that is not my answer :( – user162343 Jun 6 '15 at 15:29
• Shouldn't it be $u'=e^xu^2$ then $u=-e^{-x}$ I believe. – Ellya Jun 6 '15 at 15:38
• Ok but it has the same problem right?, because I have to use the exponential – user162343 Jun 6 '15 at 15:45
• Ah, I edited my post to solve the correct equation @user162343 – Mankind Jun 6 '15 at 15:58
• Ok thanks a lot, but then I have to solve another differential equation now substituting $u=y'$ right? – user162343 Jun 6 '15 at 16:00