Nonlinear DE with Exact Solution

$\begin{aligned} y'' + \left(\frac 1{x+e}+y'\right)y' & = 0 &\small \color{#3D99F6} \text{Let }\frac {dy}{dx} = v \implies \frac {d^2y}{dx^2} = \frac {dv}{dx} \\ \frac {dv}{dx} + \frac v{x+e} + v^2 & = 0 \\ \frac {dv}{dx} + \frac v{x+e} & = - v^2 & \small \color{#3D99F6} \text{Bernoulli's equation: }y'' + p(x)y = q(x)y^n \\ -\frac 1{v^2}\cdot \frac {dv}{dx} - \frac 1{v(x+e)} & = 1 & \small \color{#3D99F6} \text{Let }u = \frac 1v \implies \frac {du}{dx} = - \frac 1{v^2} \cdot \frac {dv}{dx} \\ \frac {du}{dx} - \frac u{x+e} & = 1 & \small \color{#3D99F6} \text{Linear differential equation: } \frac {dy}{dt} + p(t) y = g(t) \\ \frac 1{x+e} \cdot \frac {du}{dx} - \frac u{(x+e)^2} & = \frac 1{x+e} & \small \color{#3D99F6} \text{Integrating factor: } \mu(x) = e^{\int - \frac {dx}{x+e}} = \frac 1{x+e} \\ \frac d{dx} \left(\frac u{x+e}\right) & = \frac 1{x+e} \end{aligned}$

$\begin{aligned} \implies \frac u{x+e} & = \int \frac 1{x+e} dx = \ln (x+e) +\color{#3D99F6} c_1 & \small \color{#3D99F6} \text{where }c_1 \text{ is the constant of integration.} \\ u = \frac 1v = \frac {dx}{dy} & = (x+e)(\ln (x+e) + c_1) \end{aligned}$

$\begin{aligned} \implies y & = \int \frac {dx}{(x+e)(\ln (x+e) + c_1)} \\ & = \ln(\ln(x+e) + c_1)+\color{#3D99F6}c_2 & \small \color{#3D99F6} \text{where }c_2 \text{ is the constant of integration.} \\ y(0) & = \ln(1 + c_1)+c_2 = 0 & ...(1) \\ y'(x) & = \frac 1{(x+e)(\ln (x+e) + c_1)} \\ y'(0) & = \frac 1{e\color{#3D99F6}(1+c_1)} = \frac 1e & \small \color{#3D99F6} \implies c_1 = 0 \text{; putting in (1)} \\ y(0) & = \ln(1)+c_2 = 0 & \small \color{#3D99F6} \implies c_2 = 0 \\ \implies y(x) & = \ln(\ln(x+e)) \\ y(10) & = \ln(\ln(10+e)) \approx \boxed{0.933} \end{aligned}$

The equation becomes $\frac{d}{dx}\big[\ln y'(x) + \ln(x+e) + y(x)\big] \; = \; \frac{y''(x)}{y'(x)} + \frac{1}{x+e} + y'(x) \; = \; 0$ and hence $\ln y'(x) + \ln(x+e) + y(x) \; = \; c$ for some constant $c$ . Substituting $x=0$ we see that $c=0$ , and hence $(x+e)y'(x) e^{y(x)} \; = \; 1$ so that $\frac{d}{dx} e^{y(x)} \; = \; y'(x)e^{y(x)} \; = \; \frac{1}{x+e}$ and hence $e^{y(x)} \; = \; \ln(x+e) + d$ for some constant $d$ . Substituting in $x=0$ we deduce that $d=0$ , so that $y(x) \;= \; \ln(\ln(x+e))$ making the answer $\ln(\ln(10+e)) = \boxed{0.9333604015}$ .

Let $f(x) = \frac{1}{x+e}$ . Thus the equation is $y'' + fy' + y'^2 = 0$ . Let's substitute $w = y'$ to reduce the degree of the equation to get $w' + fw + w^2 = 0$ . Now, we notice that $f$ satisfies the differential equation $f' + f^2 = 0$ and in the equation we have precisely the terms $w' + w^2$ so we are being hinted at trying to exploit that property. Consider the substitution $w = zf$ . Then $w' = z'f + f'z$ . Then the equation is $z'f + u(f' + f) + z^2f^2 = 0 \implies z'f + z^2f^2 = 0 \implies z' + z^2 f = 0 \implies -z^{-2}z' = f$ .

This is now a separable equation. Integrating both sides we obtain $z^{-1} = \log(x+e) + C \implies z = \frac{1}{\log(x+e) + C}$ . Substituting back to $w$ , $w = \frac{1}{(x+e)(\log(x+e) + C)}$ . But $w=y'$ so we may apply one of the boundary conditions to find the first constant. The condition tells us $\frac{1}{e} = \frac{1}{e(1 + C)} \implies C = 0$ . Now $y' = \frac{1}{(x+e) \log(x+e)}$ . We need to compute:

$\int \frac{dx}{(x+e) \log(x+e)}$ . Let $u = \log(x+e)$ , then the integral is $\int \frac{du}{u}$ with solution $\log (u)$ . So $y = \log \log (x+e) + D$ . Applying the second boundary condition, $0 = 0 + D \implies D = 0$ .

The solution is $y = \log \log (x+e)$

The solution is $y(x) = \log(\log(x+e))$ . So, $y(10) = \log(\log(10+e)) \approx 0.93336$