Solve the differential equation d/dx(dy/dx) = e^y, that is y double prime = e^y.

Step-by-step solution

1. 1

   Recognise the equation as autonomous. The independent variable $x$ does not appear explicitly — only $y$ and its derivatives do. For such equations the standard move is to make $y$, not $x$, the independent variable, which drops the order from two to one.

2. 2

   Substitute $v=y'$ and use the chain rule in $y$. Writing $v$ as a function of $y$,

   $y''=\frac{dv}{dx}=\frac{dv}{dy}\cdot\frac{dy}{dx}=v\frac{dv}{dy}.$

   The equation becomes the first-order separable equation

   $v\frac{dv}{dy}=e^{y}.$

   This $v\,\frac{dv}{dy}$ identity is the entire technique; without it the substitution $v=y'$ leaves an equation in three variables.

3. 3

   Separate and integrate to get the first integral. From $v\,dv=e^{y}\,dy$:

   $\frac{v^{2}}{2}=e^{y}+C\;\Longrightarrow\;\left(\frac{dy}{dx}\right)^{2}=2e^{y}+C_{1}.$

   This is a conserved quantity — the mechanical analogue is energy conservation for a particle in the potential $-e^{y}$. Taking the square root,

   $\frac{dy}{dx}=\pm\sqrt{2e^{y}+C_{1}}.$

4. 4

   Separate a second time. The remaining equation is separable in $y$ and $x$:

   $\int\frac{dy}{\sqrt{2e^{y}+C_{1}}}=\pm x+C_{2}.$

   For $C_{1}=k>0$ the substitution $u=e^{y}$ turns the left side into $\int\frac{du}{u\sqrt{2u+k}}$, a standard form evaluating to

   $\frac{1}{\sqrt{k}}\ln\left|\frac{\sqrt{2e^{y}+k}-\sqrt{k}}{\sqrt{2e^{y}+k}+\sqrt{k}}\right|=\pm x+C_{2}.$

5. 5

   Invert to an explicit solution. Solving that logarithmic relation for $y$ collapses the algebra into a hyperbolic sine:

   $y=\ln\!\frac{C_{1}}{2}-2\ln\left|\sinh\!\left(\frac{\sqrt{C_{1}}\,(x-x_{0})}{2}\right)\right|,\qquad C_{1}>0,$

   where $x_{0}$ absorbs $C_{2}$. The degenerate case $C_{1}=0$ is simpler and worth quoting separately:

   $y=-2\ln\left|c-\frac{x}{\sqrt2}\right|.$

6. 6

   Verify both forms numerically. For $C_{1}=1.5$, $x_{0}=0.3$: a second-difference estimate of $y''$ at $x=1.4$ gives $1.42405$ against $e^{y}=1.42404$ ✓, and at $x=2.2$ gives $0.35950$ against $0.35950$ ✓. For the $C_{1}=0$ branch with $c=3$: at $x=1$, $y''=0.190212$ and $e^{y}=0.190210$ ✓. Both solutions blow up where the logarithm argument vanishes, which is genuine — the solution reaches infinity in finite $x$.