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Question Number 54857 by Tawa1 last updated on 13/Feb/19

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Feb/19

$$\frac{d^{2}y}{{dx}^2}+\lambda y=0$$$$y=e^{mx}$$$$m^2{e}^{mx}+\lambda e^{mx}=0$$$$e^{mx}(m^2+\lambda )=0$$$$e^{mx}\ne 0$$$$m^2=-\lambda$$$$m=\pm i\sqrt{\lambda }$$$$y={Ae}^{i\sqrt{\lambda }x}+{Be}^{-i\sqrt{\lambda }x}\left(generalsolution\right)$$$$\frac{dy}{dx}=A\left(i\sqrt{\lambda }\right)e^{i\sqrt{\lambda }x}+B(-i\sqrt{\lambda })e^{-i\sqrt{\lambda }x}$$$$puttingcondition\to {\left(\frac{dy}{dx}\right)}_{x=0}+{\left(y\right)}_{x=0}=0$$$$weget$$$$A\left(i\sqrt{\lambda }\right)+B(-i\sqrt{\lambda })+A+B=0$$$$(A+B)+i\sqrt{\lambda }(A-B)=0$$$$$$$$A+B=0A-B=0soA=0B=0$$$$peculiarconclusioniscoming...$$$$butbothA\ne 0B\ne 0$$$$somethingwrong...condition1$$$$aspercondition2$$$${Ae}^{i\sqrt{\lambda }}+{Be}^{-i\sqrt{\lambda }}+3\{A\left(i\sqrt{\lambda }\right)e^{i\sqrt{\lambda }}+B(-i\sqrt{\lambda })e^{-i\sqrt{(\lambda }}=0$$$${Ae}^{i\sqrt{\lambda }}+{Be}^{-i\sqrt{\lambda }}=0$$$$3i\sqrt{\lambda }({Ae}^{i\sqrt{\lambda }}-{Be}^{-i\sqrt{\lambda }})=0$$$${Ae}^{i\sqrt{\lambda }}={Be}^{-i\sqrt{\lambda }}$$$$samepeculiarconclusioncoming...$$$$letothercheck....$$

Commented by Tawa1 last updated on 13/Feb/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{Waiting}$$

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