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I am trying to find the solution of the equation t $y''-(\cos x) y'+(\sin x )y = 0$.

I need help urgently.Thanks

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Is there a $t$ in front of $y''$? If so is it a constant? – user17762 Jun 14 '12 at 8:42
@What is that $t$? – Nancy Rutkowskie Jun 14 '12 at 9:09
Please make sure I didn't change your question unintentionally. I don't know what's that $t$ doing there, must be a typo. – Gigili Jun 14 '12 at 9:13
someone has been researched this in another Q&A site: – doraemonpaul Jun 14 '12 at 11:40
This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date. – doraemonpaul Sep 10 '12 at 1:34

Hint: $$ -\cos x\; y'(x)+\sin x\; y(x) =(-y(x) \cos x )' $$

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This is a linear ODE of trigonometric function coefficients. The current approach of solving it is to transform it to a linear ODE of polynomial function coefficients first.

Let $u=\sin x$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=(\cos x)\dfrac{dy}{du}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left((\cos x)\dfrac{dy}{du}\right)=(\cos x)\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)-(\sin x)\dfrac{dy}{du}=(\cos x)\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}-(\sin x)\dfrac{dy}{du}=(\cos x)\dfrac{d^2y}{du^2}\cos x-(\sin x)\dfrac{dy}{du}=(\cos^2 x)\dfrac{d^2y}{du^2}-(\sin x)\dfrac{dy}{du}$

$\therefore(\cos^2 x)\dfrac{d^2y}{du^2}-(\sin x)\dfrac{dy}{du}-(\cos^2 x)\dfrac{dy}{du}+(\sin x)y=0$

$(1-\sin^2 x)\dfrac{d^2y}{du^2}+(\sin^2 x-1-\sin x)\dfrac{dy}{du}+(\sin x)y=0$


This belongs to a Heun’s confluent equation ( (, however in this case the properties are even simpler, since in this case is lucky that sum of the coefficients is equal to zero.

$\therefore y=e^u$ is a particular solution.

Let $y=e^uv$ ,

Then $\dfrac{dy}{du}=e^u\dfrac{dv}{du}+e^uv$














$y=e^{\sin x}\int\dfrac{c_2e^{-\sin x}}{\sqrt{\sin^2 x-1}}~d(\sin x)$

$y=e^{\sin x}\int\dfrac{c_2e^{-\sin x}}{i\cos x}\cos x~dx$

$y=e^{\sin x}\int C_2e^{-\sin x}~dx$

$y=e^{\sin x}\left(C_2\int_0^xe^{-\sin x}~dx+C_1\right)$

$y=C_1e^{\sin x}+C_2e^{\sin x}\int_0^xe^{-\sin x}~dx$

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