# Non-linear Second Order Differential equation

I want to ask for a hint in solving the following ODE,

$y'' (1+x^2) + y' *(x) = C$,

where C is a constant. I've tried a couple of ways to manipulate this ODE; such as letting v = y'

$v' (1+x^2) + v *(x) = C$

but I can't see a way to solve the equation for y

• Your idea is good. Note that you can divide by $1+x^2$. Then use your favorite method to solve a first degree ODE. – Git Gud May 25 '14 at 11:54
• To me it looks similar to Euler-type equation which can be simplified and solved by the substitution $x=e^t$, $t = \ln x$ (so that in case of $y''x^2 + y'x$ you get $g'' + g'$ instead). Cannot be sure it is going to work in this case. – Shady_arc May 25 '14 at 12:12

$$\left(1+x^2\right)v' + xv = C$$ we can obtain $$v\sqrt{1+x^2} = C\int\frac{1}{\sqrt{1+x^2}}dx + \lambda_{1}$$ or $$\frac{dy}{dx} = \frac{C}{\sqrt{1+x^2}}\int\frac{1}{\sqrt{1+x^2}}dx + \frac{\lambda_{1}}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+x^2}}\left[C\sinh^{-1}(x) + \lambda_1\right]$$ so $$y = \lambda_1\int \frac{1}{\sqrt{1+x^2}} + C\int \frac{1}{\sqrt{1+x^2}}\sinh^{-1}(x)$$ the first integral evaluates to $\sinh^{-1}x$ the second can be evaluated as follows $$\int \frac{1}{\sqrt{1+x^2}}\sinh^{-1}(x) = \int \sinh^{-1}(x) \frac{d}{dx}\sinh^{-1}(x) dx = \frac{1}{2}\left(\sinh^{-1}x\right)^{2} + \lambda_2$$ $$y(x) = \lambda_1 \sinh^{-1}(x) + \frac{C}{2}\left(\sinh^{-1}x\right)^{2} + \lambda_3$$