Question Number 15179 by Joel577 last updated on 08/Jun/17
$$\mathrm{If}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{4}{y}\:=\:\mathrm{0}\: \\ $$$${y}\left(\mathrm{0}\right)\:=\:\mathrm{1}\:\mathrm{and}\:{y}\left(\frac{\pi}{\mathrm{6}}\right)\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:{y}\left({x}\right)\:? \\ $$
Answered by sma3l2996 last updated on 08/Jun/17
$${y}''\left({x}\right)+\mathrm{4}{y}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} +\mathrm{4}=\mathrm{0}\Leftrightarrow{r}_{\mathrm{1}} =\mathrm{2}{i}\:,\:\:{r}=−\mathrm{2}{i} \\ $$$${y}\left({x}\right)={Ae}^{\mathrm{2}{ix}} +{Be}^{−\mathrm{2}{ix}} \\ $$$${y}\left(\mathrm{0}\right)={A}+{B}=\mathrm{1}\Leftrightarrow{B}=\mathrm{1}−{A} \\ $$$${y}\left(\frac{\pi}{\mathrm{6}}\right)={Ae}^{\frac{\pi}{\mathrm{3}}{i}} +\left(\mathrm{1}−{A}\right){e}^{−\frac{\pi}{\mathrm{3}}{i}} ={A}\left({e}^{\frac{\pi}{\mathrm{3}}{i}} −{e}^{−\frac{\pi}{\mathrm{3}}{i}} \right)+\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\mathrm{2}{iAsin}\left(\frac{\pi}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{iA}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{2}{iA}={i}+\mathrm{1}\Leftrightarrow{A}=\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$${y}\left({x}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({e}^{{i}\left(\mathrm{2}{x}−\frac{\pi}{\mathrm{4}}\right)} +{e}^{−{i}\left(\mathrm{2}{x}−\frac{\pi}{\mathrm{4}}\right)} \right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}×\mathrm{2}{cos}\left(\mathrm{2}{x}−\frac{\pi}{\mathrm{4}}\right) \\ $$$${y}\left({x}\right)=\sqrt{\mathrm{2}}{cos}\left(\mathrm{2}{x}−\frac{\pi}{\mathrm{4}}\right) \\ $$
Commented by Joel577 last updated on 08/Jun/17
$${thank}\:{you}\:{very}\:{much} \\ $$