Solve $y'' + e^y = 0$

Key concept: This nonlinear differential equation can be reduced by using a standard substitution. Since $y$ depends on $x$, we treat $y''$ as the second derivative with respect to $x$ and integrate carefully.

Step by step

We solve

$y''+e^y=0.$

A useful first step is to multiply both sides by $y'$:

$y'y''+e^y y'=0.$

Now recognize each term as a derivative:

$y'y''=\frac{d}{dx}\left(\frac{(y')^2}{2}\right), \qquad e^y y'=\frac{d}{dx}(e^y).$

So the equation becomes

$\frac{d}{dx}\left(\frac{(y')^2}{2}+e^y\right)=0.$

Integrating once gives

$\frac{(y')^2}{2}+e^y=C_1.$

Hence

$(y')^2=2(C_1-e^y).$

Taking square roots,

$y'=\pm\sqrt{2(C_1-e^y)}.$

Now separate variables:

$\frac{dy}{\sqrt{2(C_1-e^y)}}=\pm dx.$

This implicit integral gives the general solution in quadrature form:

$\int \frac{dy}{\sqrt{2(C_1-e^y)}}=\pm x+C_2.$

So the solution is generally expressed implicitly rather than by an elementary closed form.

Pitfall alert

A frequent mistake is to try to integrate $y''+e^y=0$ as if it were linear. It is nonlinear because of the $e^y$ term. Another common error is forgetting to multiply by $y'$ before integrating, which is what creates the conserved-energy form.

Try different conditions

If the equation were $y''+e^x=0$, then it would be linear and easy to integrate twice. If an initial condition such as $y(0)$ and $y'(0)$ were provided, you could determine the constants $C_1$ and $C_2$ and write a fully specified implicit solution.

Further reading

nonlinear differential equation, first integral, separation of variables