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Inserting BC; \( u(0)=0 \) We find that we must choose \( c_{1} \) such that \[ \int_{a_{1}}^{0} f(y) \sin k y d y=0 \Rightarrow c_{1}=0 \] Hence (4.2.6) reduces to \[ u(x)=\frac{\cos k x}{k} \int_{0}^{x} f(y) \sin k y d y-\frac{\sin k x}{k} \int_{c_{2}}^{x} f(y) \cos k y d y \] which imply we must choose \( c_{1}=0 \) Using second BC in (4.2.8) we have \[ \int_{c_{2}}^{l} f(y) \cos k y d y=\frac{\cos k l}{\sin k l} \int_{0}^{l} f(y) \sin k y d y \] After slight manipulation we can re write the above equation as \[ \begin{array}{l} \int_{c_{2}}^{0} f(y) \cos k y d y=\frac{1}{\sin k l} \int_{0}^{l} f(y)[\sin k y \cos k l-\cos k y \sin k l] d y \\ \quad \Rightarrow \int_{c_{2}}^{0} f(y) \cos k y d y=\frac{1}{\sin k l} \int_{0}^{l} f(y)[\sin k(y-l)] d y \end{array} \] \( \therefore \) Solution can now be written in the form \[ \begin{aligned} u(x)= & \frac{\cos k x}{k} \int_{0}^{x} f(y) \sin k y d y-\frac{\sin k x}{k} \int_{0}^{x} f(y) \cos k y d y \\ & -\frac{\sin k x}{k \sin k l} \int_{0}^{l} f(y) \sin k(y-l) d y \end{aligned} \]

          Inserting BC; \( u(0)=0 \)
We find that we must choose \( c_{1} \) such that
\[
\int_{a_{1}}^{0} f(y) \sin k y d y=0 \Rightarrow c_{1}=0
\]

Hence (4.2.6) reduces to
\[
u(x)=\frac{\cos k x}{k} \int_{0}^{x} f(y) \sin k y d y-\frac{\sin k x}{k} \int_{c_{2}}^{x} f(y) \cos k y d y
\]
which imply we must choose \( c_{1}=0 \)
Using second BC in (4.2.8) we have
\[
\int_{c_{2}}^{l} f(y) \cos k y d y=\frac{\cos k l}{\sin k l} \int_{0}^{l} f(y) \sin k y d y
\]

After slight manipulation we can re write the above equation as
\[
\begin{array}{l}
\int_{c_{2}}^{0} f(y) \cos k y d y=\frac{1}{\sin k l} \int_{0}^{l} f(y)[\sin k y \cos k l-\cos k y \sin k l] d y \\
\quad \Rightarrow \int_{c_{2}}^{0} f(y) \cos k y d y=\frac{1}{\sin k l} \int_{0}^{l} f(y)[\sin k(y-l)] d y
\end{array}
\]
\( \therefore \) Solution can now be written in the form
\[
\begin{aligned}
u(x)= & \frac{\cos k x}{k} \int_{0}^{x} f(y) \sin k y d y-\frac{\sin k x}{k} \int_{0}^{x} f(y) \cos k y d y \\
& -\frac{\sin k x}{k \sin k l} \int_{0}^{l} f(y) \sin k(y-l) d y
\end{aligned}
\]

Inserting BC; u(0)=0
We find that we must choose c1 such that

∫a1^0 f(y) sin k y d y=0 ⇒ c1=0

Hence (4.2.6) reduces to

u(x)=(cos k x)/(k)∫0^x f(y) sin k y d y-(sin k x)/(k)∫c2^x f(y) cos k y d y

which imply we must choose c1=0
Using second BC in (4.2.8) we have

∫c2^l f(y) cos k y d y=(cos k l)/(sin k l)∫0^l f(y) sin k y d y

After slight manipulation we can re write the above equation as

∫c2^0 f(y) cos k y d y=(1)/(sin k l)∫0^l f(y)[sin k y cos k l-cos k y sin k l] d y
⇒∫c2^0 f(y) cos k y d y=(1)/(sin k l)∫0^l f(y)[sin k(y-l)] d y

∴ Solution can now be written in the form

u(x)= (cos k x)/(k)∫0^x f(y) sin k y d y-(sin k x)/(k)∫0^x f(y) cos k y d y
-(sin k x)/(k sin k l)∫0^l f(y) sin k(y-l) d y

Added by Martin A.

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