Solving Integral Equation for y(\pi/3)

[asmani]

Homework Statement

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

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

Homework Equations

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?

Thanks in advance.

 

[jackmell]

Homework Statement

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

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

Homework Equations

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?

Thanks in advance.

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

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.

 

[uart]

since y(1)=1

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

 

[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?

 

[jackmell]

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

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

(forgot the negative sign)

 

[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.

 

[HallsofIvy]

Oh, blast! I completely neglected the exponential part!

Thanks.

 

[gabbagabbahey]

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

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

 

[uart]

The initial condition y(1)=1 comes from the original form of the question.
y(1)=\text{e}^{-\int_{1}^{1} \frac{ y(t) }{\sin^2 (t)} dt } = \text{e}^{0} = 1

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.

 

[asmani]

Thanks anybody.