2
$\begingroup$

I have this here ODE:

$$xy'' = y' + x((y')^2 + x^2)$$

Naturally, I'd try this substitution first: $$y' = p, p=p(x)$$

The equation then transforms into $$xp' = p+x(p^2+x^2)$$ Dividing it by $x$, I get

$$p' = \frac{p}{x} + p^2 + x^2$$

Which is a Riccati equation with the solution: $$p = x \cdot \tan{(\frac{x^2}{2}+C_1)}$$

The thing is, if I substitute back $y'=p$, the integral on the right side is not an easy one to solve, and even if I do solve it with WolframAlpha the solutions are not the same as if I plug in the second-order equation directly. It makes me wonder if I should have tried another substitution/method.

Any help will be appreciated!

$\endgroup$
2
  • $\begingroup$ @LutzLehmann thanks fixed $\endgroup$
    – john doe
    May 17, 2021 at 14:43
  • $\begingroup$ The equation is of the first order in $y'$. $\endgroup$
    – user65203
    May 17, 2021 at 14:45

3 Answers 3

3
$\begingroup$

$$y' = x \cdot \tan{\left(\frac{x^2}{2}+C\right)}$$ $$\dfrac {dy}{du}\dfrac {du}{dx} = x \cdot \tan(u)$$ $$\dfrac {dy}{du} = \tan(u)$$ Where $u=\dfrac {x^2}{2}+C$ then integrate. $$y=-\ln |\cos u |+K$$

$\endgroup$
1
$\begingroup$

You have now a formula of the form $$ y'(x)=f(u(x))u'(x) $$ with $f(u)=\tan(u)$ and $u(x)=\frac{x^2}2+c$. This the gives that $$ y(x)=F(u(x))+d, $$ where $F'=f$, here $F(u)=-\ln|\cos(u)|$

$\endgroup$
1
$\begingroup$

Let $y'=xz$. The equation becomes

$$xz+x^2z'=xz+x(x^2z^2+x^2),$$ which is separable:

$$\frac{z'}{z^2+1}=x$$ or $$\arctan z=\frac{x^2}2+c$$ and $$y'=x\tan\left(\frac{x^2}2+c\right)$$ as you found.


Now the integral is not difficult: $$\int x\tan\left(\frac{x^2}2+c\right)dx=\int \tan\left(u+c\right)du=-\log(\cos(u+c))+c'\\ =-\log\left(\cos\left(\frac{x^2}2+c\right)\right)+c'.$$

$\endgroup$

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.