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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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2  
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: tw.knowledge.yahoo.com/question/question?qid=1011101609747 –  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

2 Answers 2

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

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

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

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

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

This belongs to a Heun’s confluent equation (http://dlmf.nist.gov/31.12) (http://www.maplesoft.com/support/help/Maple/view.aspx?path=HeunC), 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 $\dfrac{dy}{du}=e^u\dfrac{dv}{du}+e^uv$

$\dfrac{d^2y}{du^2}=e^u\dfrac{d^2v}{du^2}+e^u\dfrac{dv}{du}+e^u\dfrac{dv}{du}+e^uv=e^u\dfrac{d^2v}{du^2}+2e^u\dfrac{dv}{du}+e^uv$

$\therefore(1-u^2)\left(e^u\dfrac{d^2v}{du^2}+2e^u\dfrac{dv}{du}+e^uv\right)+(u^2-u-1)\left(e^u\dfrac{dv}{du}+e^uv\right)+ue^uv=0$

$(1-u^2)\left(\dfrac{d^2v}{du^2}+2\dfrac{dv}{du}+v\right)+(u^2-u-1)\left(\dfrac{dv}{du}+v\right)+uv=0$

$(1-u^2)\dfrac{d^2v}{du^2}+2(1-u^2)\dfrac{dv}{du}+(1-u^2)v+(u^2-u-1)\dfrac{dv}{du}+(u^2-u-1)v+uv=0$

$(1-u^2)\dfrac{d^2v}{du^2}-(u^2+u-1)\dfrac{dv}{du}=0$

$(u^2-1)\dfrac{d^2v}{du^2}=-(u^2+u-1)\dfrac{dv}{du}$

$\dfrac{\dfrac{d^2v}{du^2}}{\dfrac{dv}{du}}=-1-\dfrac{u}{u^2-1}$

$\int\dfrac{\dfrac{d^2v}{du^2}}{\dfrac{dv}{du}}~du=\int\left(-1-\dfrac{u}{u^2-1}\right)~du$

$\ln\dfrac{dv}{du}=-u-\dfrac{1}{2}\ln(u^2-1)+c_1$

$\dfrac{dv}{du}=\dfrac{c_2e^{-u}}{\sqrt{u^2-1}}$

$v=\int\dfrac{c_2e^{-u}}{\sqrt{u^2-1}}~du$

$ye^{-u}=\int\dfrac{c_2e^{-u}}{\sqrt{u^2-1}}~du$

$y=e^u\int\dfrac{c_2e^{-u}}{\sqrt{u^2-1}}~du$

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

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

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

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

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

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