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So usually, a differential equation question is asking to find a general solution. But this is the other way around.

I have a general solution $$y=\frac{1}{c_1 \cos x+c_2 \sin x},$$ and I want to find the differential equation to it. This, I think, is about finding $c_1$ and $c_2$. So, I calculated the derivative, $$y'=\frac{c_1 \sin x -c_1 \cos x}{(c_2 \sin x+c_2 \cos x)^2}.$$

Now, it's time to subtract $y-y'$ and let them cancel out to find $c_1$,$c_2$ right? Or is the next step to find $y''$ and see if they have cancelling out terms and find $c_1$ and $c_2$?

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    $\begingroup$ Since you have two parameters, one would expect that you need $y$, $y'$, and $y''$ if you want to do this one directly. $\endgroup$ Sep 16 at 17:21
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    $\begingroup$ $c_1\cos x+c_2\sin x$ is known to be the general solution of $y''+y=0$. Hence, $$\left(\dfrac1y\right)''+\dfrac1y=0.$$ $\endgroup$
    – user958916
    Sep 16 at 17:42
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    $\begingroup$ Is there a general approach for y=f(x)? Or, must we rely on insight and experience? $\endgroup$ Sep 17 at 17:06
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$$y=\frac{1}{c_1 \cos x+c_2 \sin x}$$ $$\dfrac 1 y={c_1 \cos x+c_2 \sin x}$$ Substitute $u=1/y$: $$u={c_1 \cos x+c_2 \sin x}$$ $$u''+u=0$$ It's easier now..

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    $\begingroup$ yes i understand now. thank you $\endgroup$ Sep 17 at 0:12
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Ode solution of the reciprocal of y can be recognized, not repeated.

Primes are differentiation w.r.t $x$

$$\dfrac 1 y={c_1 \cos x+c_2 \sin x}$$ $$\left(\frac{1}{y}\right)'' + \left(\frac{1}{y}\right)=0 $$ $$\left(\frac{y^2y''-2y y^{'2}}{y^4}\right) + \left(\frac{1}{y}\right)=0 $$ $$ y y''-2 y^{'2}+y^2=0. $$

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$$\frac 1{y\cos x}=c_1+c_2\tan x\to-\frac{y'\cos x-y\sin x}{y^2\cos^2x}=\frac{c_2}{\cos^2\theta}\iff-\frac{y'\cos x-y\sin x}{y^2}=c_2.$$ So differentiating again the numerator yields $$(y''\cos x-2y'\sin x+y\cos x)y^2-2(y'\cos x-y\sin x)yy'=0,$$ or after simplification $$y''y-2y'^2+y^2=0.$$


This is a general technique that often works: isolate one of the constants as a term and differentiate. That makes it vanish.

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