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Question Number 15179 by Joel577 last updated on 08/Jun/17

If (d^2 y/dx^2 ) + 4y = 0 y(0) = 1 and y((π/6)) = ((√3)/2) + (1/2) How to find y(x) ?

$$\mathrm{If}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{4}{y}\:=\:\mathrm{0}\: \\ $$$${y}\left(\mathrm{0}\right)\:=\:\mathrm{1}\:\mathrm{and}\:{y}\left(\frac{\pi}{\mathrm{6}}\right)\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:{y}\left({x}\right)\:? \\ $$

Answered by sma3l2996 last updated on 08/Jun/17

y′′(x)+4y=0 r^2 +4=0⇔r_1 =2i , r=−2i y(x)=Ae^(2ix) +Be^(−2ix) y(0)=A+B=1⇔B=1−A y((π/6))=Ae^((π/3)i) +(1−A)e^(−(π/3)i) =A(e^((π/3)i) −e^(−(π/3)i) )+(1/2)−i((√3)/2) =2iAsin((π/3))+(1/2)−i((√3)/2)=((√3)/2)+(1/2) 2iA((√3)/2)+(1/2)−i((√3)/2)−((√3)/2)−(1/2)=0 2iA=i+1⇔A=(1/2)−i(1/2)=((√2)/2)e^(−i(π/4)) B=(1/2)+i(1/2)=((√2)/2)e^(i(π/4)) y(x)=((√2)/2)(e^(i(2x−(π/4))) +e^(−i(2x−(π/4))) )=((√2)/2)×2cos(2x−(π/4)) y(x)=(√2)cos(2x−(π/4))

$${y}''\left({x}\right)+\mathrm{4}{y}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} +\mathrm{4}=\mathrm{0}\Leftrightarrow{r}_{\mathrm{1}} =\mathrm{2}{i}\:,\:\:{r}=−\mathrm{2}{i} \\ $$$${y}\left({x}\right)={Ae}^{\mathrm{2}{ix}} +{Be}^{−\mathrm{2}{ix}} \\ $$$${y}\left(\mathrm{0}\right)={A}+{B}=\mathrm{1}\Leftrightarrow{B}=\mathrm{1}−{A} \\ $$$${y}\left(\frac{\pi}{\mathrm{6}}\right)={Ae}^{\frac{\pi}{\mathrm{3}}{i}} +\left(\mathrm{1}−{A}\right){e}^{−\frac{\pi}{\mathrm{3}}{i}} ={A}\left({e}^{\frac{\pi}{\mathrm{3}}{i}} −{e}^{−\frac{\pi}{\mathrm{3}}{i}} \right)+\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\mathrm{2}{iAsin}\left(\frac{\pi}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{iA}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{2}{iA}={i}+\mathrm{1}\Leftrightarrow{A}=\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$${y}\left({x}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({e}^{{i}\left(\mathrm{2}{x}−\frac{\pi}{\mathrm{4}}\right)} +{e}^{−{i}\left(\mathrm{2}{x}−\frac{\pi}{\mathrm{4}}\right)} \right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}×\mathrm{2}{cos}\left(\mathrm{2}{x}−\frac{\pi}{\mathrm{4}}\right) \\ $$$${y}\left({x}\right)=\sqrt{\mathrm{2}}{cos}\left(\mathrm{2}{x}−\frac{\pi}{\mathrm{4}}\right) \\ $$

Commented by Joel577 last updated on 08/Jun/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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