A differential equation that you might have or might not have seen before

Let $\frac{dy}{dx} = g(x)$ $\Rightarrow$ $\frac{d^2y}{dx^2}=\frac{dg}{dx}$ , we can now rewrite the given equation as:

$\frac{dg}{dx} + g^2 = 1$ $\Leftrightarrow$ $\frac{dg}{dx}=1-g^2$

$\Leftrightarrow$ $dx=\frac{dg}{1-g^2}$

$\Leftrightarrow$ $\int{dx} = x = \int{\frac{1}{1-g^2}dg} = \frac{1}{2}ln|\frac{1+g}{1-g}| + C_1$

Since $x=0$ when $\frac{dy}{dx}=g=0$ , $C_1 = 0$

$\Rightarrow$ $x=\frac{1}{2}|\frac{1+g}{1-g}|$ $\Leftrightarrow$ $e^{2x}=\frac{1+g}{1-g}$

$\Rightarrow$ $e^{2x}(1-g)=1+g$

$\Rightarrow$ $g=\frac{e^{2x}-1}{e^{2x}+1}$

or $\frac{dy}{dx}=\frac{e^{2x}-1}{e^{2x}+1}$

Thus, we have:

$\int{dy} = y = \int{\frac{e^{2x}-1}{e^{2x}+1}dx}$

Let $t=e^{2x}$ $\Rightarrow$ $dt=2e^{2x}dx=2tdx$

$\Rightarrow$ $y= \frac{1}{2}\int{\frac{t-1}{t(t+1)}dt}=\int{\frac{1}{t+1}dt} - \frac{1}{2}\int{\frac{1}{t}dt}$

$\Rightarrow$ $y=ln|t+1|-\frac{1}{2}ln|t| + C_2$

$\Rightarrow$ $y=ln(1+e^{2x}) -x + C_2$

$y=0$ when $x=0$

$\Rightarrow$ $C_2 = -ln2$

$\Rightarrow$ $y= ln(1+e^{2x}) -x -ln2 = \boxed{-x + ln[\frac{1}{2}(1+e^{2x})]}$

Similar solution as @Russell Ngo 's.

$\begin{aligned} \frac {d^2y}{dx^2} + \left(\frac {dy}{dx}\right)^2 & = 1 & \small \color{#3D99F6} \text{Let }u = \frac {dy}{dx} \implies \frac {du}{dx} = \frac {d^2y}{dx^2} \\ \frac {du}{dx} + u^2 & = 1 \\ \frac {du}{1-u^2} & = dx \\ \int \frac {du}{(1-u)(1+u)} & = \int dx \\ \int \frac 12 \left(\frac 1{1-u}+\frac 1{1+u}\right) du & = \int dx \\ -\ln(1-u) + \ln(1+u) & = 2x + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ \ln \left(\frac {1+u}{1-u} \right) & = 2x & \small \color{#3D99F6} \text{Since }u(0) = \frac {dy}{dx}\bigg|_{x=0} = 0 \implies C = 0 \\ \frac {1+u}{1-u} & = e^{2x} & \small \color{#3D99F6} \text{Put back }u = \frac {dy}{dx} \text{ and rearrange.} \\ (1+e^{2x})\frac {dy}{dx} & = e^{2x} - 1 \end{aligned}$

$\begin{aligned} \implies y & = \int \frac {e^{2x}-1}{e^{2x}+1} dx \\ & = \frac 12 \ln (1+e^{2x}) -\color{#3D99F6} \int \frac 1{1+e^{2x}} dx & \small \color{#3D99F6} \text{Let }t = e^x \implies dt = e^x \ dx \\ & = \frac 12 \ln (1+e^{2x}) -\color{#3D99F6} \int \frac 1{t(1+t^2)} dt \\ & = \frac 12 \ln (1+e^{2x}) - \int \left(\frac 1t - \frac t{1+t^2}\right) dt \\ & = \frac 12 \ln (1+e^{2x}) - \ln t + \frac 12 \ln (1+t^2) + C_1 & \small \color{#3D99F6} \text{Put back }t = e^x \\ & = \ln (1+e^{2x}) - x + C_1 & \small \color{#3D99F6} \text{Since }y(0) = 0 \\ 0 & = \ln 2 - 0 + C_1 & \small \color{#3D99F6} \implies C_1 = - \ln 2 \\ \implies y & = \ln (1+e^{2x}) - x - \ln 2 \\ & = \boxed{\ln \left(\dfrac {1+e^{2x}}2\right) - x} \end{aligned}$