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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 :) )

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

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

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

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