2
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Your idea is good. Note that you can divide by $1+x^2$. Then use your favorite method to solve a first degree ODE. $\endgroup$ – Git Gud May 25 '14 at 11:54
  • $\begingroup$ 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. $\endgroup$ – Shady_arc May 25 '14 at 12:12
1
$\begingroup$

$$ \left(1+x^2\right)v' + xv = C $$ 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 $$ \int \frac{1}{\sqrt{1+x^2}}\sinh^{-1}(x) = \int \sinh^{-1}(x) \frac{d}{dx}\sinh^{-1}(x) dx = \frac{1}{2}\left(\sinh^{-1}x\right)^{2} + \lambda_2 $$ $$ y(x) = \lambda_1 \sinh^{-1}(x) + \frac{C}{2}\left(\sinh^{-1}x\right)^{2} + \lambda_3 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.