Find the general solution of y'' + (y')^2 = 2e^(-y)

Find the general solution of $y'' + (y')^2 = 2e^{-y}$.

Step 1: Substitution

Let $p = y'$. Express $y''$ using the chain rule: $y'' = p\,\frac{dp}{dy}.$

The equation becomes: \[p\,\frac{dp}{dy} + p^2 = 2e^{-y}. \tag{1}\]

Step 2: Linearize with $u = p^2$

Let $u = p^2 = (y')^2$, so $\dfrac{du}{dy} = 2p\dfrac{dp}{dy}$. Multiply (1) by 2: \[\frac{du}{dy} + 2u = 4e^{-y}. \tag{2}\]

This is a first-order linear ODE in $u(y)$.

Step 3: Solve the linear ODE

The integrating factor is $\mu = e^{2y}$. Multiply (2) by $e^{2y}$: $\frac{d}{dy}\left(e^{2y}u\right) = 4e^{y}.$

Integrate: $e^{2y}u = 4e^y + C_1,\qquad u = 4e^{-y} + C_1 e^{-2y}.$

Substitute back $u = (y')^2$: \[(y')^2 = e^{-2y}(4e^y + C_1). \tag{3}\]

Step 4: Second integration

Take the square root and separate variables: $\frac{e^y\,dy}{\sqrt{4e^y + C_1}} = \pm\,dx.$

Use $t = e^y$, $dt = e^y\,dy$: $\frac{dt}{\sqrt{4t + C_1}} = \pm\,dx.$

Integrate using $s = 4t+C_1$: \[\frac{\sqrt{4e^y + C_1}}{2} = \pm(x + C_2),\qquad 4e^y + C_1 = 4(x+C_2)^2.\]

Solve for $y$ (relabelling $C_1/4 \to C_1$): $e^y = (x+C_2)^2 - C_1.$

Step 5: General solution

The solution is recorded in the result below.

Step 6: Verification

Denote $Q = (x+C_2)^2 - C_1$, so $y = \ln Q$. Then: \[y' = \frac{2(x+C_2)}{Q},\qquad y'' = \frac{2Q - 4(x+C_2)^2}{Q^2} = \frac{-2(x+C_2)^2-2C_1}{Q^2}.\]

Check $y'' + (y')^2$: \[\frac{-2(x+C_2)^2-2C_1}{Q^2} + \frac{4(x+C_2)^2}{Q^2} = \frac{2(x+C_2)^2-2C_1}{Q^2} = \frac{2Q}{Q^2} = \frac{2}{Q} = 2e^{-y}.\ \checkmark\]