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Integral equation

$y(x) = 1 + \lambda\int\limits_0^2\cos(x-t) y(t) \mathrm{d}t$ has:

  1. a unique solution for $\lambda \neq \frac{4}{\pi +2}$;

  2. a unique solution for $\lambda \neq \frac{4}{\pi -2}$;

  3. no solution for $\lambda \neq \frac{4}{\pi +2}$, but the corresponding homogeneous equation has a non-trivial solution; or

  4. no solution for $\lambda \neq \frac{4}{\pi -2}$, but the corresponding homogeneous equation has a non-trivial solution.

I am stuck on this problem. Can anyone help me please?

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    If it is a question of some Indian Entrance Examination please add the source (Exam. name, year)in the title of this question. It will be helpful to other students using this site.2014-01-21

2 Answers 2

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Hint: Use $\cos(x-t)=\cos x\cos t+\sin x\sin t$ and derive twice. After that you get: $y^{\prime\prime}(x)=-(\lambda\int_{0}^2\cos(x-t)y(t)dt)$ and so $y^{\prime\prime}+y-1=0$ The corresponding homogenous equation is $y^{\prime\prime}+y=0$ The solution is $y=a\cos x+b\sin x+1$. We just have to subsititute it back in the original equation. Then, $a\cos x+b\sin x+1=1+\lambda \int_{0}^2\cos(x-t)(a\cos t+b\sin t+1)dt$ and so $a\cos x+b\sin x=\lambda \cos x\int_{0}^2\cos t(a\cos t+b\sin t+1)dt+\\\lambda \sin x\int_{0}^2\sin t(a\cos t+b\sin t+1)dt$ Therefore, we have the system of equations: \begin{gather}a=\lambda \int_{0}^2\cos t(a\cos t+b\sin t+1)dt\\ b=\lambda \int_{0}^2\sin t(a\cos t+b\sin t+1)dt\end{gather}

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    OKK its somewhat lengthy..........2012-12-18
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Related technique: (I), (II). The integral equation you have is a "Fredholm equation of the 2nd kind with seperable kernel". There are standard techniques to solve this type of equations. See here, page 20 for the method and a worked example how to find such $\lambda$.

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    This is indeed a *Fredholm equation of the 2nd kind with seperable kernel*... but the fact that is irrelevant. Please try not to send the OP on deadends.2012-12-25