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

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. –  Dutta Jan 21 at 11:06

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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$a\doteq \lambda \sin t \int_0^2 (a\cos t \sin t + b\sin^2 t + \sin t) dt$.it getting hard –  pankaj Dec 18 '12 at 16:36
its getting hard ,i have to then solve for homogeneous also,is there any shortcut........ –  pankaj Dec 18 '12 at 16:48
guide me sir............. –  pankaj Dec 18 '12 at 17:00
@pankaj I don't know any shortcut you can take other than computing the integrals, that are by the way easier than you think. Look at Mhenni's answer for more info –  Nameless Dec 18 '12 at 17:02
@pankaj It is. As a sidenote, if you plan to post more questions at MSE, you may want to increase your accept rate and get more people interested –  Nameless Dec 18 '12 at 17:13
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 $\lambda$.