Boundary Value Problem asked by David Helton

Exercise:

Solve the following Boundary Value Problem , $y''+3y=0\,\,,$ $y\left( 0 \right)=7\,\,\,\,,\,\,\,\,y\left( 2\pi  \right)=0$

Solution: by using the characteristic equation we get ${{r}^{2}}+3=0$ How ??

Simply Take $y\left( t \right)={{e}^{rt}}\Leftrightarrow y'\left( t \right)=r{{e}^{rt}}\,\,\And \,\,y''\left( t \right)={{r}^{2}}{{e}^{rt}}$

Substitute to get ${{r}^{2}}{{e}^{rt}}+3{{e}^{rt}}=0\Rightarrow {{e}^{rt}}\left( {{r}^{2}}+3 \right)=0$ but ${{e}^{rt}}>0$ so ${{r}^{2}}+3=0$

$\Rightarrow {{r}^{2}}=-3\Rightarrow r=\pm \sqrt{3{{i}^{2}}}\Rightarrow r=\pm i\sqrt{3}$ So $y\left( t \right)={{c}_{1}}\,\cos \left( \sqrt{3}t \right)+{{c}_{2}}\sin \left( \sqrt{3}t \right)$

$y\left( 0 \right)=7\Leftrightarrow 7={{c}_{1}}\cos \left( 0 \right)+{{c}_{2}}\sin \left( 0 \right)\Leftrightarrow 7={{c}_{1}}$

$y\left( 2\pi  \right)=0\Leftrightarrow 0={{c}_{1}}\cos \left( 2\pi \sqrt{3} \right)+{{c}_{2}}\sin \left( 2\pi \sqrt{3} \right)$

So $7\cos \left( 2\pi \sqrt{3} \right)+{{c}_{2}}\sin \left( 2\pi \sqrt{3} \right)=0\Leftrightarrow {{c}_{2}}=-\frac{7\cos \left( 2\pi \sqrt{3} \right)}{\sin \left( 2\pi \sqrt{3} \right)}=-7\cot \left( 2\pi \sqrt{3} \right)$

Hence $y\left( t \right)=7\cos \left( \sqrt{3}t \right)-7\cot \left( 2\pi \sqrt{3} \right)\sin \left( \sqrt{3}t \right)$