I was supposed to find the differential equation of the equation $$(x-h)^2+(y-k)^2=a^2$$ where $a$ is a given constant.

I did it the normal way by differentiating once to get an equation in terms of $(x-h)$ and $(y-k)$. Then differentiating it once more to get $(y-k)$ in terms of $y'$ and $y''$. Then substituting it in the first equation to get $(x-k)$ in terms of $y'$ and $y''$. Then substituting it in the original equation. Finally, I got the differential equation as $$[1+(y')^2]^3=a^2y''^2$$

However, I thought of another simpler way. Assume $$x-h=a\:cos\theta,\;\;y-k=a\:sin\theta$$ $$\frac{dx}{d\theta}=-a\:sin\theta,\;\;\frac{dy}{d\theta}=a\:cos\theta$$ $$\frac{d^2x}{d\theta^2}=-a\:cos\theta,\;\;\frac{d^2y}{d\theta^2}=-a\:sin\theta$$ $$\frac{dy}{dx}=-cot\theta,\;\;\frac{d^2y}{dx^2}=tan\theta$$

Hence, I get the differential equation as $$y'y''+1=0$$

Is my second method valid? Since I'm supposed to solve such types of sums in less than $3$ minutes, I find second method a lot easier.

I would also like to know if there's some other simple method to find the differential equation of such equations.

EDIT: I have uploaded the steps to the first method. enter image description here


There is a mistake in your second method. $\frac{d^2 y}{dx^2}$ is not equal to $\tan \theta$. It's more complicated.

With the first method, your result $[1+(y')^2]^3=a^2y''$ is not correct : try to solve this ODE and see that the solution is not $(x-h)^2+(y-k)^2=a^2$ as expected. Since you didn't published the details of the calculus, it is not possible to see where is the mistake.

My result is : $$[1+(y')^2]^{3/2}=-a y''$$

  • $\begingroup$ Please see the edit. I didn't find anything wrong in that. Still, I'll try once to solve the ODE. Though to my misfortune, I don't have enough expertise to solve second degree differential equations. $\endgroup$ – Tejas Mar 22 '15 at 6:26
  • $\begingroup$ I agree with your new edit : $$[1+(y')^2]^3=a^2y''^2$$ In your first answer (before the edit), you forgot the power $2$ of $y''^2$. Now, it is consistent with my result $$[1+(y')^2]^{3/2}=-ay''$$ in selecting the convenient sign of the square root. $\endgroup$ – JJacquelin Mar 22 '15 at 7:10
  • $\begingroup$ Corrected it. Thanks for noticing! Can you explain why the second method fails, and if I were to follow the second method, what differential equation would I end up with? $\endgroup$ – Tejas Mar 22 '15 at 10:20
  • $\begingroup$ As already said, the second method fails because your second derivative $\frac{d^2y}{dx^2}=\tan(\theta)$ is false. With the correct expression $\frac{d^2y}{dx^2}=-\frac{1}{a\sin^3(\theta)}$ one end up with the same differential equation than the first method. $\endgroup$ – JJacquelin Mar 22 '15 at 10:44
  • $\begingroup$ Use the chain rule : $$\frac{d^2y}{dx^2}=\left(\frac{d}{d\theta}(-\cot\theta)\right)\frac{d\theta}{dx}$$ $\endgroup$ – JJacquelin Mar 22 '15 at 11:00

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