your favorite method

logistic_guy

logistic_guy

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Solve the differential equation by your favorite method to get Green's function.

\(\displaystyle \frac{d^2y}{dx^2} + k^2y = f(x)\)

\(\displaystyle a < x < b\)
 
💜

\(\displaystyle A'(x)\cos kx + B'(x)\sin kx = 0\)

\(\displaystyle -kA'(x)\sin kx + kB'(x)\cos kx = f(x)\)
 
logistic_guy said:
\(\displaystyle A'(x)= -\frac{B'(x)\sin kx}{\cos kx}\)
Click to expand...
We will plug this result in the second equation and we will solve for \(\displaystyle B'(x)\).

\(\displaystyle -kA'(x)\sin kx + kB'(x)\cos kx = f(x)\)


\(\displaystyle -k\left(-\frac{B'(x)\sin kx}{\cos kx}\right)\sin kx + kB'(x)\cos kx = f(x)\)


\(\displaystyle kB'(x)\sin^2 kx + kB'(x)\cos^2 kx = f(x)\cos kx\)


\(\displaystyle kB'(x)(\sin^2 kx + \cos^2 kx) = f(x)\cos kx\)


\(\displaystyle kB'(x)(1) = f(x)\cos kx\)


\(\displaystyle kB'(x) = f(x)\cos kx\)


\(\displaystyle B'(x) = \frac{f(x)\cos kx}{k}\)

Then,

\(\displaystyle A'(x)= -\frac{f(x)\sin kx}{k}\)
 
logistic_guy said:
\(\displaystyle \frac{d^2y}{dx^2} + k^2y = f(x)\)
Click to expand...
From the variation of parameters method, we know that the general solution to this differential equation is:

\(\displaystyle y(x) = A(x)\cos kx + B(x)\sin kx\)

From the previous post, we get:

\(\displaystyle y(x) = -\frac{\cos kx}{k}\int_{\alpha}^{x}f(s)\sin ks \ ds + \frac{\sin kx}{k} \int_{\beta}^{x}f(s)\cos ks \ ds\)

Or

\(\displaystyle y(x) = \frac{\sin kx}{k} \int_{\beta}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{\alpha}^{x}f(s)\sin ks \ ds\)

where the constants \(\displaystyle \alpha\) and \(\displaystyle \beta\) can be found from the boundary conditions!
 
Now to get Green's function, we should have the following boundary conditions:

\(\displaystyle y(a) = y(b) = 0\)

Let us apply the first boundary condition.

\(\displaystyle 0 = \frac{\sin ka}{k} \int_{\beta}^{a}f(s)\cos ks \ ds - \frac{\cos ka}{k}\int_{\alpha}^{a}f(s)\sin ks \ ds\)


\(\displaystyle \frac{\cos ka}{k}\int_{\alpha}^{a}f(s)\sin ks \ ds = \frac{\sin ka}{k} \int_{\beta}^{a}f(s)\cos ks \ ds\)


\(\displaystyle \int_{\alpha}^{a}f(s)\sin ks \ ds = \frac{\sin ka}{\cos ka} \int_{\beta}^{a}f(s)\cos ks \ ds \ \ \ \ \textcolor{red}{\bold{(1)}}\)


At this moment, we don't know how to get rid of \(\displaystyle \alpha\) and \(\displaystyle \beta\). We will try something. It may fail. If it did fail, we would try to think of something else! We will try to do this trick:

\(\displaystyle \int_{\alpha}^{x} = \int_{\alpha}^{a} + \int_{a}^{x}\)

Back to our solution:

\(\displaystyle y(x) = \frac{\sin kx}{k} \int_{\beta}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{\alpha}^{x}f(s)\sin ks \ ds\)


\(\displaystyle y(x) = \frac{\sin kx}{k} \int_{\beta}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\left(\int_{\alpha}^{a}f(s)\sin ks \ ds + \int_{a}^{x}f(s)\sin ks \ ds\right)\)

Now we will replace the first term in the brackets by \(\displaystyle \textcolor{red}{\bold{(1)}}\)

\(\displaystyle y(x) = \frac{\sin kx}{k} \int_{\beta}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\left(\frac{\sin ka}{\cos ka} \int_{\beta}^{a}f(s)\cos ks \ ds + \int_{a}^{x}f(s)\sin ks \ ds\right)\)

Simplify.

\(\displaystyle y(x) = \frac{\sin kx}{k} \int_{\beta}^{x}f(s)\cos ks \ ds - \frac{\cos kx\sin ka}{k\cos ka} \int_{\beta}^{a}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)


We got rid of \(\displaystyle \alpha\)😍🥳🤩
 
Let us combine the \(\displaystyle \beta\) in one integral before we apply the second boundary condition.


\(\displaystyle y(x) = \frac{\sin kx}{k} \int_{\beta}^{x}f(s)\cos ks \ ds - \frac{\cos kx\sin ka}{k\cos ka} \int_{\beta}^{a}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)



\(\displaystyle y(x) = \frac{\sin kx}{k} \left(\int_{\beta}^{a}f(s)\cos ks \ ds + \int_{a}^{x}f(s)\cos ks \ ds\right) - \frac{\cos kx\sin ka}{k\cos ka} \int_{\beta}^{a}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)



\(\displaystyle y(x) = \frac{\sin kx}{k}\int_{\beta}^{a}f(s)\cos ks \ ds + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx\sin ka}{k\cos ka} \int_{\beta}^{a}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)



\(\displaystyle y(x) = \left(\frac{\sin kx}{k} - \frac{\cos kx\sin ka}{k\cos ka}\right)\int_{\beta}^{a}f(s)\cos ks \ ds + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)
 
Apply the second boundary condition.

\(\displaystyle 0 = \left(\frac{\sin kb}{k} - \frac{\cos kb\sin ka}{k\cos ka}\right)\int_{\beta}^{a}f(s)\cos ks \ ds + \frac{\sin kb}{k}\int_{a}^{b}f(s)\cos ks \ ds - \frac{\cos kb}{k}\int_{a}^{b}f(s)\sin ks \ ds\)



\(\displaystyle \left(\frac{\cos kb\sin ka}{k\cos ka} - \frac{\sin kb}{k}\right)\int_{\beta}^{a}f(s)\cos ks \ ds = \frac{\sin kb}{k}\int_{a}^{b}f(s)\cos ks \ ds - \frac{\cos kb}{k}\int_{a}^{b}f(s)\sin ks \ ds\)


Let us simplify \(\displaystyle \left(\frac{\cos kb\sin ka}{k\cos ka} - \frac{\sin kb}{k}\right)\).


\(\displaystyle \left(\frac{\cos kb\sin ka}{k\cos ka} - \frac{\sin kb}{k}\right) = \left(\frac{\cos kb\sin ka}{k\cos ka} - \frac{\sin kb \cos ka}{k\cos ka}\right) = \frac{\sin k(a - b)}{k\cos ka}\)

This gives us:

\(\displaystyle \frac{\sin k(a - b)}{k\cos ka}\int_{\beta}^{a}f(s)\cos ks \ ds = \frac{\sin kb}{k}\int_{a}^{b}f(s)\cos ks \ ds - \frac{\cos kb}{k}\int_{a}^{b}f(s)\sin ks \ ds\)

Or

\(\displaystyle \int_{\beta}^{a}f(s)\cos ks \ ds = \frac{k\cos ka}{\sin k(a - b)}\left(\frac{\sin kb}{k}\int_{a}^{b}f(s)\cos ks \ ds - \frac{\cos kb}{k}\int_{a}^{b}f(s)\sin ks \ ds\right)\)

Or

\(\displaystyle \int_{\beta}^{a}f(s)\cos ks \ ds = \frac{\cos ka}{\sin k(a - b)}\left(\sin kb\int_{a}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{b}f(s)\sin ks \ ds\right)\)
 
Back to our solution.

\(\displaystyle y(x) = \left(\frac{\sin kx}{k} - \frac{\cos kx\sin ka}{k\cos ka}\right)\int_{\beta}^{a}f(s)\cos ks \ ds + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)

Substitute the last result we obtained.

\(\displaystyle y(x) = \left(\frac{\sin kx}{k} - \frac{\cos kx\sin ka}{k\cos ka}\right)\left[\frac{\cos ka}{\sin k(a - b)}\left(\sin kb\int_{a}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{b}f(s)\sin ks \ ds\right) \right] + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)

The expression (solution) looks very complicated and I still don't know if my method works.

🤣😛

Let us try to simplify and see what happens.

Let us start by simplifying the brackets: \(\displaystyle \left(\frac{\sin kx}{k} - \frac{\cos kx\sin ka}{k\cos ka}\right)\)

\(\displaystyle \left(\frac{\sin kx}{k} - \frac{\cos kx\sin ka}{k\cos ka}\right) = \left(\frac{\sin kx\cos ka}{k\cos ka} - \frac{\cos kx\sin ka}{k\cos ka}\right) = \frac{\sin k(x - a)}{k\cos ka}\)

This gives:

\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\cos ka}\left[\frac{\cos ka}{\sin k(a - b)}\left(\sin kb\int_{a}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{b}f(s)\sin ks \ ds\right) \right] + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)



\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\sin k(a - b)}\left(\sin kb\int_{a}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{b}f(s)\sin ks \ ds\right) + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)



\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\sin k(a - b)}\left(\sin kb\int_{a}^{x}f(s)\cos ks \ ds + \sin kb\int_{x}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{x}f(s)\sin ks \ ds - \cos kb\int_{x}^{b}f(s)\sin ks\right) + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)



\(\displaystyle y(x) = \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds + \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\cos ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin ks + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)



\(\displaystyle y(x) = \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds + \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\cos ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin ks \ ds + \frac{\sin kx\sin ka\cos kb - \sin kx\sin kb\cos ka}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx\sin ka\cos kb - \cos kx\sin kb\cos ka}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds\)



\(\displaystyle y(x) = \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds + \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\cos ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin ks \ ds - \frac{\sin kx\sin kb\cos ka}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds + \frac{\sin kx\sin ka\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx\sin ka\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds + \frac{\cos kx\sin kb\cos ka}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds\)

Ahhhhhh that's tough!

😭😭

@khansaheb

Can you check where I went wrong?
🤔
 
logistic_guy said:
Back to our solution.

\(\displaystyle y(x) = \left(\frac{\sin kx}{k} - \frac{\cos kx\sin ka}{k\cos ka}\right)\int_{\beta}^{a}f(s)\cos ks \ ds + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)

Substitute the last result we obtained.

\(\displaystyle y(x) = \left(\frac{\sin kx}{k} - \frac{\cos kx\sin ka}{k\cos ka}\right)\left[\frac{\cos ka}{\sin k(a - b)}\left(\sin kb\int_{a}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{b}f(s)\sin ks \ ds\right) \right] + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)

The expression (solution) looks very complicated and I still don't know if my method works.

🤣😛

Let us try to simplify and see what happens.

Let us start by simplifying the brackets: \(\displaystyle \left(\frac{\sin kx}{k} - \frac{\cos kx\sin ka}{k\cos ka}\right)\)

\(\displaystyle \left(\frac{\sin kx}{k} - \frac{\cos kx\sin ka}{k\cos ka}\right) = \left(\frac{\sin kx\cos ka}{k\cos ka} - \frac{\cos kx\sin ka}{k\cos ka}\right) = \frac{\sin k(x - a)}{k\cos ka}\)

This gives:

\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\cos ka}\left[\frac{\cos ka}{\sin k(a - b)}\left(\sin kb\int_{a}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{b}f(s)\sin ks \ ds\right) \right] + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)



\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\sin k(a - b)}\left(\sin kb\int_{a}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{b}f(s)\sin ks \ ds\right) + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)



\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\sin k(a - b)}\left(\sin kb\int_{a}^{x}f(s)\cos ks \ ds + \sin kb\int_{x}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{x}f(s)\sin ks \ ds - \cos kb\int_{x}^{b}f(s)\sin ks\right) + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)



\(\displaystyle y(x) = \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds + \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\cos ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin ks + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)



\(\displaystyle y(x) = \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds + \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\cos ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin ks \ ds + \frac{\sin kx\sin ka\cos kb - \sin kx\sin kb\cos ka}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx\sin ka\cos kb - \cos kx\sin kb\cos ka}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds\)



\(\displaystyle y(x) = \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds + \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\cos ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin ks \ ds - \frac{\sin kx\sin kb\cos ka}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds + \frac{\sin kx\sin ka\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx\sin ka\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds + \frac{\cos kx\sin kb\cos ka}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds\)

Ahhhhhh that's tough!

😭😭

@khansaheb

Can you check where I went wrong?
🤔
Click to expand...
No.....
 
khansaheb said:
No.....
Click to expand...
Check whether the DE is satisfied by the "solution" through actually differentiating each term.
Then evaluate each term at the given boundaries and check those values against the given values of BCs.
If everything matches - you are home free....
 
khansaheb said:
Check whether the DE is satisfied by the "solution" through actually differentiating each term.
Then evaluate each term at the given boundaries and check those values against the given values of BCs.
If everything matches - you are home free....
Click to expand...
No. Either you solve it and show the audience where I went wrong, or we will ban you temporary as you have been useless during the last \(\displaystyle 7\) months. It's your call!

😏😏
 
We will try to use a different approach now. We will start at this part:

\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\sin k(a - b)}\left(\sin kb\int_{a}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{b}f(s)\sin ks \ ds\right) + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)

This time we will start simplifying what inside the brackets.

\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\sin k(a - b)}\left(\int_{a}^{b}f(s)\sin k(b - s) \ ds\right) + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)
 
We will combine the second and third terms together.

\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\sin k(a - b)}\left(\int_{a}^{b}f(s)\sin k(b - s) \ ds\right) + \frac{1}{k}\int_{a}^{x}f(s)\sin k(x - s) \ ds\)
 
Multiply the second term and divide it by \(\displaystyle \sin k(a - b)\)

\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\sin k(a - b)}\left(\int_{a}^{b}f(s)\sin k(b - s) \ ds\right) + \frac{\sin k(a - b)}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin k(x - s) \ ds\)
 
Let us do this trick again:

\(\displaystyle \int_{a}^{b} = \int_{a}^{x} + \int_{x}^{b}\)

\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\sin k(a - b)}\left(\int_{a}^{x}f(s)\sin k(b - s) \ ds + \int_{x}^{b}f(s)\sin k(b - s) \ ds\right) + \frac{\sin k(a - b)}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin k(x - s) \ ds\)

\(\displaystyle \textcolor{red}{\bold{simplify.}}\)

\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin k(b - s) \ ds + \frac{\sin k(a - b)}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin k(x - s) \ ds + \frac{\sin k(x - a)}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin k(b - s) \ ds\)

\(\displaystyle \textcolor{red}{\bold{Or}}\)

\(\displaystyle y(x) = \frac{1}{k\sin k(a - b)}\left(\sin k(x - a)\int_{a}^{x}f(s)\sin k(b - s) \ ds + \sin k(a - b)\int_{a}^{x}f(s)\sin k(x - s) \ ds\right) + \frac{\sin k(x - a)}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin k(b - s) \ ds\)

We need to find a way to combine \(\displaystyle \int_{a}^{x}\) together to construct Green's function.
 
logistic_guy said:
We need to find a way to combine \(\displaystyle \int_{a}^{x}\) together to construct Green's function.
Click to expand...
\(\displaystyle \sin A\sin B = \frac{1}{2}\bigg[\cos(A - B) - \cos(A + B)\bigg]\)

Then,

\(\displaystyle \sin k(x - a)\sin k(b - s) = \frac{1}{2}\bigg[\cos k(x - a - b + s) - \cos k(x - a + b - s)\bigg]\)
And
\(\displaystyle \sin k(a - b)\sin k(x - s) = \frac{1}{2}\bigg[\cos k(a - b - x + s) - \cos k(a - b + x - s)\bigg]\)

If we sum these together and use the identity \(\displaystyle \cos(-\theta) = \cos(\theta)\), we get:

\(\displaystyle \sin k(x - a)\sin k(b - s) + \sin k(a - b)\sin k(x - s) = \frac{1}{2}\bigg[\cos k(x - a - b + s) - \cos k(a - b + x - s)\bigg]\)


\(\displaystyle = \frac{1}{2}\bigg[\cos k[ \ (x - b) - (a - s) \ ] - \cos k[ \ (x - b) + (a - s) \ ]\bigg] = \sin k(x - b)\sin k(a - s)\)

\(\displaystyle \textcolor{red}{\bold{Then,}}\)

\(\displaystyle y(x) = \frac{1}{k\sin k(a - b)}\left(\sin k(x - a)\int_{a}^{x}f(s)\sin k(b - s) \ ds + \sin k(a - b)\int_{a}^{x}f(s)\sin k(x - s) \ ds\right) + \frac{\sin k(x - a)}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin k(b - s) \ ds\)

\(\displaystyle \textcolor{red}{\bold{Or}}\)

\(\displaystyle y(x) = \frac{1}{k\sin k(a - b)}\left(\int_{a}^{x} f(s)\bigg[\sin k(x - a)\sin k(b - s) + \sin k(a - b)\sin k(x - s)\bigg] ds\right) + \frac{\sin k(x - a)}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin k(b - s) \ ds\)

\(\displaystyle \textcolor{red}{\bold{Or}}\)

\(\displaystyle y(x) = \frac{1}{k\sin k(a - b)}\left(\int_{a}^{x} f(s)\bigg[\sin k(x - b)\sin k(a - s)\bigg] ds\right) + \frac{\sin k(x - a)}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin k(b - s) \ ds\)

\(\displaystyle \textcolor{red}{\bold{Or}}\)

\(\displaystyle y(x) = \frac{\sin k(x - b)}{k\sin k(a - b)}\int_{a}^{x} f(s)\sin k(a - s) \ ds + \frac{\sin k(x - a)}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin k(b - s) \ ds\)

\(\displaystyle \textcolor{red}{\bold{Or}}\)

\(\displaystyle y(x) = \int_{a}^{b} f(s)g(x,s) \ ds\)

where \(\displaystyle g(x,s)\) is the Green's function and it is defined as:

[imath]\large g(x,s) =\begin{cases} \frac{\sin k(a - s)\sin k(x - b)}{k\sin k(a - b)} & a \leq s < x\\[2ex] \frac{\sin k(x - a)\sin k(b - s)}{k\sin k(a - b)} & x < s \leq b\end{cases}[/imath]

\(\displaystyle \textcolor{green}{\bold{Or}}\)

[imath]\large g(x,s) =\begin{cases} -\frac{\sin k(s - a)\sin k(b - x)}{k\sin k(b-a)} & a \leq s < x\\[2ex] -\frac{\sin k(x - a)\sin k(b - s)}{k\sin k(b-a)} & x < s \leq b\end{cases}[/imath]
 
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