3
$\begingroup$

I have a differential equation $$xy''(x) +(n+1-x)y'(x) + ay(x)=0.$$ If I set $x=r^t$ then how to plug in this and how to use change of variable to get the differential equation for $r$ instead of $x,$ i.e. the following equation:

$$\frac{dy}{dx}=\frac{dy}{dr}\frac{dr}{dx}$$

$\endgroup$
  • $\begingroup$ You sure that last term is $ay'(x)$ and not just $ay(x)$? $\endgroup$ – Kaster Jun 18 '13 at 16:15
  • $\begingroup$ Sorry it's a typo, it should be y(x) $\endgroup$ – Keith Jun 18 '13 at 16:16
  • $\begingroup$ Anyway, you have a good start. As for the sub I'd recommend to find $y_x$ through $y_r = y_x x_r$, not if that matters, just might cause less confusion. $\endgroup$ – Kaster Jun 18 '13 at 16:20
9
$\begingroup$

By chain rule $$\frac{dy}{dx}=\frac{dy}{dr}\frac{dr}{dx}$$ $$\frac{d^2y}{dx^2}=\frac{d^2y}{dr^2}\bigg(\frac{dr}{dx}\bigg)^2+\frac{dy}{dr}\frac{d^2r}{dx^2}$$ where $$x=r^t\Rightarrow dx=tr^{t-1}dr\Rightarrow\frac{dr}{dx}=\frac 1{tr^{t-1}}$$ $$dx^2=t(t-1)r^{t-2}dr^2\Rightarrow\frac{d^2r}{dx^2}=\frac 1{t(t-1)r^{t-2}}$$ By plugging into original equation $$r^t \Bigg(\frac{d^2y}{dr^2}\bigg(\frac 1{tr^{t-1}}\bigg)^2+\frac{dy}{dr}\frac 1{t(t-1)r^{t-2}} \Bigg)+(n+1-r^t)\frac{dy}{dr}\frac 1{tr^{t-1}} + ay=0$$

---- Addition for chain rule ---- $$\frac{d}{dx}\bigg(\frac{dy}{dx}\bigg)=\frac{d}{dx}\bigg(\frac{dy}{dr}\frac{dr}{dx}\bigg)$$ $$\frac{d^2y}{dx^2}=\frac{d}{dx}\bigg(\frac{dy}{dr}\bigg)\frac{dr}{dx}+\frac{dy}{dr}\frac{d}{dx}\bigg(\frac{dr}{dx}\bigg)$$ $$\frac{d^2y}{dx^2}=\frac{d^2y}{dr^2}\frac{dr}{dx}\frac{dr}{dx}+\frac{dy}{dr}\frac{d^2r}{dx^2}$$

$\endgroup$
  • $\begingroup$ how does $\frac{d^2y}{dx^2}= \frac{d^2y}{dr^2}\bigg(\frac{dr}{dx}\bigg)^2+\frac{dy}{dr}\frac{d^2r}{dx^2}$ follow from the chain rule? $\endgroup$ – Michael Angelo Mar 26 '16 at 15:22
  • 1
    $\begingroup$ @fawningflagellum Please check the added section $\endgroup$ – AnilB Mar 28 '16 at 11:09
  • $\begingroup$ @MichaelAngelo By applying $\frac{d}{dx}=(\frac{dr}{dx})\frac{d}{dr}$ to both sides of the chain rule equation above, and using the product rule on the right hand side. $\endgroup$ – Travis Bemrose Aug 9 '18 at 19:16

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.