Let $\phi$ satisfy $\phi(x)=f(x)+\int_0^x\sin(x-t)\phi(t)\,dt$. Then $\phi$ is given by?

Question

Let $\phi$ satisfy,

$$\phi(x)=f(x)+\int_0^x\sin(x-t)\phi(t)\,dt$$

Then $\phi$ is given by,

$\phi(x)=f(x)+\int_0^x(x-t)\phi(t)\,dt$
$\phi(x)=f(x)+\int_0^x\sin(x-t)\phi(t)\,dt$
$\phi(x)=f(x)+\int_0^x\cos(x-t)\phi(t)\,dt$
$\phi(x)=f(x)-\int_0^x\sin(x-t)\phi(t)\,dt$

This question is already asked but I am not clear with that answer and I am a new contributor to stack exchange so I don't have 50 ruputations to post comment in that place. so can anyone please tell me how to solve this...I tried but I get $\phi''(x)=f''(x)$

If you have a question about a previous Q&A then it would be helpful to add a link to that question, and explain your doubts about the answer. — Martin R, Jun 09 '19 at 17:54
ss a start differentiate all equations by $x$ then sub in 1 to 5 to see what works. — pshmath0, Jun 09 '19 at 18:04
In the current version of the question, isn't (2) exactly the given equation? — JimmyK4542, Jun 10 '19 at 00:17
Possible duplicate of Which of the following satisfies the equation $\phi(x)=f(x)+\int_{0}^{x}\sin(x-t)\phi(t)dt$ — Jesse P Francis, Jul 26 '19 at 13:24

score 2 · Answer 1 · answered Jun 09 '19 at 18:09

2

You have$$\phi(x)=f(x)+\sin(x)*\phi(x)$$Let the Laplace transform of $\phi(t),\mathcal L[\phi(t)]=H(s)$ and that of $f(t)$ be $F(s)$. Take the Laplace transform of both sides:$$H(s)=F(s)+\frac{H(s)}{s^2+1}\\\implies H(s)=\left(1+\frac1{s^2}\right)F(s)$$Taking the inverse Laplace transform,$$\phi(t)=f(t)+\int_0^t(t-\tau)f(\tau)d\tau,t\ge0$$You can also write$$\phi(t)=f(t)+\int_0^tf(t-\tau)\tau~d\tau,t\ge0$$

answered Jun 09 '19 at 18:09

Shubham Johri

17,659

Sorry sir...Can you please explain the problem without using laplace transform – Gopinath Jun 10 '19 at 00:20

Let $\phi$ satisfy $\phi(x)=f(x)+\int_0^x\sin(x-t)\phi(t)\,dt$. Then $\phi$ is given by?

1 Answers1