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Question Number 47624 by Umar last updated on 12/Nov/18
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{d}.\mathrm{e}\:\mathrm{using}\:\mathrm{method}\:\mathrm{of}\:\mathrm{variation} \\ $$$$\mathrm{of}\:\mathrm{parameter}. \\ $$$$ \\ $$$$\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }+\mathrm{3}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{2y}=\mathrm{sin}\left(\mathrm{e}^{\mathrm{x}} \right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18
![let t=e^x t=e^x (dt/dx)=e^x =t (dy/dx)=(dy/dt)×(dt/dx)=t(dy/dt) (d/dx)((dy/dx))=(d/dt)(t(dy/dt))×(dt/dx) =t×{t(d^2 y/dt^2 )+(dy/dt)}=t^2 (d^2 y/dt^2 )+t(dy/dt) now (d^2 y/dx^2 )+3(dy/dx)+2y=sin(e^x ) t^2 (d^2 y/dt^2 )+t(dy/dt)+3t(dy/dt)+2y=sin(t) t^2 (d^2 y/dt^2 )+4t(dy/dt)+2y=sint now [θ(θ−1)+4θ+2]y=sint θ=(d/dt) for C.F θ^2 −θ+4θ+2=0 (θ+1)(θ+2)=0 θ=−1,−2 so C.F=C_1 e^(−t) +C_2 e^(−2t) →C_1 e^(−e^x ) +C_2 e^(−2e^x ) P.I y=(1/(θ^2 +3θ+2))×sint =((θ^2 +2−3θ)/((θ^2 +2+3θ)(θ^2 +2−3θ)))×sint =((θ^2 +2−3θ)/((θ^2 +2)^2 −9θ^2 ))sint =(1/((−1^2 +2)^2 −9(−1^2 )))×{−sint+2sint−cost} =(1/(1+9))×{sint−cost} =(1/(10))×{sin(e^x )−cos(e^x )} so complete solution is C_1 e^(−e^x ) +C_2 e^(−2e^x ) +(1/(10)){sin(e^x )−cos(e^x )} pls check...](Q47631.png)
$${let} \\ $$$${t}={e}^{{x}} \\ $$$${t}={e}^{{x}} \\ $$$$\frac{{dt}}{{dx}}={e}^{{x}} ={t} \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{dt}}×\frac{{dt}}{{dx}}={t}\frac{{dy}}{{dt}} \\ $$$$\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right)=\frac{{d}}{{dt}}\left({t}\frac{{dy}}{{dt}}\right)×\frac{{dt}}{{dx}} \\ $$$$\:\:={t}×\left\{{t}\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }+\frac{{dy}}{{dt}}\right\}={t}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }+{t}\frac{{dy}}{{dt}} \\ $$$${now} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{3}\frac{{dy}}{{dx}}+\mathrm{2}{y}={sin}\left({e}^{{x}} \right) \\ $$$${t}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }+{t}\frac{{dy}}{{dt}}+\mathrm{3}{t}\frac{{dy}}{{dt}}+\mathrm{2}{y}={sin}\left({t}\right) \\ $$$${t}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }+\mathrm{4}{t}\frac{{dy}}{{dt}}+\mathrm{2}{y}={sint} \\ $$$${now} \\ $$$$\left[\theta\left(\theta−\mathrm{1}\right)+\mathrm{4}\theta+\mathrm{2}\right]{y}={sint}\:\:\:\:\theta=\frac{{d}}{{dt}} \\ $$$${for} \\ $$$${C}.{F} \\ $$$$\theta^{\mathrm{2}} −\theta+\mathrm{4}\theta+\mathrm{2}=\mathrm{0} \\ $$$$\left(\theta+\mathrm{1}\right)\left(\theta+\mathrm{2}\right)=\mathrm{0} \\ $$$$\theta=−\mathrm{1},−\mathrm{2} \\ $$$${so}\:{C}.{F}={C}_{\mathrm{1}} {e}^{−{t}} +{C}_{\mathrm{2}} {e}^{−\mathrm{2}{t}} \rightarrow{C}_{\mathrm{1}} {e}^{−{e}^{{x}} } +{C}_{\mathrm{2}} {e}^{−\mathrm{2}{e}^{{x}} } \\ $$$${P}.{I} \\ $$$${y}=\frac{\mathrm{1}}{\theta^{\mathrm{2}} +\mathrm{3}\theta+\mathrm{2}}×{sint} \\ $$$$=\frac{\theta^{\mathrm{2}} +\mathrm{2}−\mathrm{3}\theta}{\left(\theta^{\mathrm{2}} +\mathrm{2}+\mathrm{3}\theta\right)\left(\theta^{\mathrm{2}} +\mathrm{2}−\mathrm{3}\theta\right)}×{sint} \\ $$$$=\frac{\theta^{\mathrm{2}} +\mathrm{2}−\mathrm{3}\theta}{\left(\theta^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} −\mathrm{9}\theta^{\mathrm{2}} }{sint} \\ $$$$=\frac{\mathrm{1}}{\left(−\mathrm{1}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} −\mathrm{9}\left(−\mathrm{1}^{\mathrm{2}} \right)}×\left\{−{sint}+\mathrm{2}{sint}−{cost}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{9}}×\left\{{sint}−{cost}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}×\left\{{sin}\left({e}^{{x}} \right)−{cos}\left({e}^{{x}} \right)\right\} \\ $$$${so}\:{complete}\:{solution}\:{is} \\ $$$${C}_{\mathrm{1}} \:{e}^{−{e}^{{x}} } +{C}_{\mathrm{2}} {e}^{−\mathrm{2}{e}^{{x}} } +\frac{\mathrm{1}}{\mathrm{10}}\left\{{sin}\left({e}^{{x}} \right)−{cos}\left({e}^{{x}} \right)\right\} \\ $$$${pls}\:{check}... \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Umar last updated on 12/Nov/18
$$\mathrm{thanks} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18
$${most}\:{welcome}... \\ $$