Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

2 Answers 2

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}

share|improve this answer
    
$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
1  
@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
show 4 more comments

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 $\lambda$.

share|improve this answer
    
its hard ,i have to solve for homogeneous and non homogeneous also,is there any shortcut..... –  pankaj Dec 18 '12 at 16:50
    
@pankaj: Where did this problem come from? –  Mhenni Benghorbal Dec 18 '12 at 17:13
    
@pankaj: From what course has this problem come from? –  Mhenni Benghorbal Dec 19 '12 at 14:55
    
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. –  Did Dec 25 '12 at 19:08
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.