# Homework Help: An integral equation

1. Aug 1, 2012

### asmani

1. The problem statement, all variables and given/known data

Find $y(\pi/3)$ in the following integral equation:

$$y(x)=e^{-\int_{1}^{x}\frac{y(t)}{\sin^2(t)}dt}$$
2. Relevant equations

3. The attempt at a solution

Differentiating both sides gives a differential equation with the general solution $y(x)=(-\cot(x)+c)^{-1}$, and since y(1)=1, we have $c=1+\cot(1)$.

Here is two questions:

1. The hint says the answer is $y(\pi/3)=3/(6-\sqrt{3})$. Is this correct?

2. Is there any shortcut way to solve this problem without differentiating?

2. Aug 1, 2012

### jackmell

No,no. You can check it you know. Here's how in Mathematica:

Code (Text):

In[20]:=
y[x_] := 1/(-Cot[x] + Cot[1] + 1)
tval = Pi/3;
Exp[-NIntegrate[y[t]/Sin[t]^2, {t, 1, tval}]]
N[3/(6 - Sqrt[3])]
N[y[\[Pi]/3]]

Out[22]= 0.939194

Out[23]= 0.702914

Out[24]= 0.939194

at least that's what I'm turin' in.

Last edited: Aug 1, 2012
3. Aug 1, 2012

### uart

You wont get the given answer with that initial condition. Are you sure it's not $y(\pi/4) = 1\,$?

4. Aug 1, 2012

### HallsofIvy

Differentiating both sides of the integral equation gives
$$\frac{dy}{dt}= \frac{y}{sin^2(t)}$$
which separates as
$$\frac{dy}{y}= \frac{dt}{sin^2(t)}= cosec^2(t)dt$$

That integrates as
$$ln(y)= - cot(t)+ C$$
Did you accidently write $y^2$ instead of y?

5. Aug 1, 2012

### jackmell

I don't think so Hall. I belive it's

$$y'=-\frac{y^2}{\sin^2(x)}$$

(forgot the negative sign)

Last edited: Aug 1, 2012
6. Aug 1, 2012

### uart

I took logs of both sides first and then took derivatives, (using the chain rule on the LHS and the FTC on the RHS), to verify the OPs proposed solution was indeed correct.

7. Aug 1, 2012

### HallsofIvy

Oh, blast! I completely neglected the exponential part!

Thanks.

8. Aug 2, 2012

### gabbagabbahey

The initial condition $y(1)=1$ comes from the original form of the question. So either there is a typo in the question, or the given answer:

$$y(1)=\text{e}^{-\int_{1}^{1} \frac{ y(t) }{\sin^2 (t)} dt } = \text{e}^{0} = 1$$

@asmani, your solution looks fine to me

9. Aug 3, 2012

### uart

Arh yes I see. It's a definite integral so y(1)=1 is not an arbitrary initial condition.

In that case then the answer in the book is definitely wrong.

10. Aug 4, 2012

### asmani

Thanks anybody.