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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

(d^2 y/dx^2 )+λy=0  y=e^(mx)   m^2 e^(mx) +λe^(mx) =0  e^(mx) (m^2 +λ)=0  e^(mx) ≠0  m^2 =−λ  m=±i(√λ)   y=Ae^(i(√λ) x) +Be^(−i(√λ) x)     (general solution)  (dy/dx)=A(i(√λ) )e^(i(√λ) x) +B(−i(√λ) )e^(−i(√λ) x)   putting condition→((dy/dx))_(x=0) +(y)_(x=0) =0  we get  A(i(√λ) )+B(−i(√λ) )+A+B=0  (A+B)+i(√λ) (A−B)=0    A+B=0  A−B=0 so A=0  B=0  peculiar conclusion is coming...  but both A≠0   B≠0  something wrong...condition 1  as per condition 2  Ae^(i(√λ) ) +Be^(−i(√λ) ) +3{A(i(√λ))e^(i(√λ)) +B(−i(√λ) )e^(−i(√((λ)) ) =0  Ae^(i(√λ)) +Be^(−i(√λ) ) =0  3i(√λ) (Ae^(i(√λ) ) −Be^(−i(√λ) ) )=0  Ae^(i(√λ) ) =Be^(−i(√λ) )   same peculiar conclusioncoming...  let other check....

$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\lambda{y}=\mathrm{0} \\ $$$${y}={e}^{{mx}} \\ $$$${m}^{\mathrm{2}} {e}^{{mx}} +\lambda{e}^{{mx}} =\mathrm{0} \\ $$$${e}^{{mx}} \left({m}^{\mathrm{2}} +\lambda\right)=\mathrm{0} \\ $$$${e}^{{mx}} \neq\mathrm{0} \\ $$$${m}^{\mathrm{2}} =−\lambda \\ $$$${m}=\pm{i}\sqrt{\lambda}\: \\ $$$${y}={Ae}^{{i}\sqrt{\lambda}\:{x}} +{Be}^{−{i}\sqrt{\lambda}\:{x}} \:\:\:\:\left({general}\:{solution}\right) \\ $$$$\frac{{dy}}{{dx}}={A}\left({i}\sqrt{\lambda}\:\right){e}^{{i}\sqrt{\lambda}\:{x}} +{B}\left(−{i}\sqrt{\lambda}\:\right){e}^{−{i}\sqrt{\lambda}\:{x}} \\ $$$${putting}\:{condition}\rightarrow\left(\frac{{dy}}{{dx}}\right)_{{x}=\mathrm{0}} +\left({y}\right)_{{x}=\mathrm{0}} =\mathrm{0} \\ $$$${we}\:{get} \\ $$$${A}\left({i}\sqrt{\lambda}\:\right)+{B}\left(−{i}\sqrt{\lambda}\:\right)+{A}+{B}=\mathrm{0} \\ $$$$\left({A}+{B}\right)+{i}\sqrt{\lambda}\:\left({A}−{B}\right)=\mathrm{0} \\ $$$$ \\ $$$${A}+{B}=\mathrm{0}\:\:{A}−{B}=\mathrm{0}\:{so}\:{A}=\mathrm{0}\:\:{B}=\mathrm{0} \\ $$$${peculiar}\:{conclusion}\:{is}\:{coming}... \\ $$$${but}\:{both}\:{A}\neq\mathrm{0}\:\:\:{B}\neq\mathrm{0} \\ $$$${something}\:{wrong}...{condition}\:\mathrm{1} \\ $$$${as}\:{per}\:{condition}\:\mathrm{2} \\ $$$${Ae}^{{i}\sqrt{\lambda}\:} +{Be}^{−{i}\sqrt{\lambda}\:} +\mathrm{3}\left\{{A}\left({i}\sqrt{\lambda}\right){e}^{{i}\sqrt{\lambda}} +{B}\left(−{i}\sqrt{\lambda}\:\right){e}^{−{i}\sqrt{\left(\lambda\right.}\:} =\mathrm{0}\right. \\ $$$${Ae}^{{i}\sqrt{\lambda}} +{Be}^{−{i}\sqrt{\lambda}\:} =\mathrm{0} \\ $$$$\mathrm{3}{i}\sqrt{\lambda}\:\left({Ae}^{{i}\sqrt{\lambda}\:} −{Be}^{−{i}\sqrt{\lambda}\:} \right)=\mathrm{0} \\ $$$${Ae}^{{i}\sqrt{\lambda}\:} ={Be}^{−{i}\sqrt{\lambda}\:} \\ $$$${same}\:{peculiar}\:{conclusioncoming}... \\ $$$${let}\:{other}\:{check}.... \\ $$

Commented by Tawa1 last updated on 13/Feb/19

God bless you sir. Waiting

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

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