Question Number 60921 by necx1 last updated on 27/May/19
$${x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:{x}\frac{{dy}}{{dx}}\:+\:{y}=\mathrm{0} \\ $$$${please}\:{solve}\:{this}\:{Euler}\:{equation} \\ $$
Answered by tanmay last updated on 27/May/19
![x=e^t →(dx/dt)=e^t [t=lnx] (dy/dx)=(dy/dt)×(dt/dx)=(1/e^t )×(dy/dt) (dy/dt)=e^t ×(dy/dx)→x(dy/dx) (d/dx)((dy/dx)) =(d/dt)((dy/dx))×(dt/dx) =(d/dt)((1/e^t )×(dy/dt))×(1/e^t ) (1/e^t )×[(1/e^t )×(d^2 y/dt^2 )+e^(−t) ×−1×(dy/dt)] =((1/e^t ))^2 ×(d^2 y/dt^2 )−((1/e^t ))^2 ×(dy/dt) so (d^2 y/dx^2 )=((1/e^t ))^2 [(d^2 y/dt^2 )−(dy/dt)] x^2 ×(d^2 y/dx^2 )=(d^2 y/dt^2 )−(dy/dt) so eq^n x^2 (d^2 y/dx^2 )+x(dy/dx)+y=0 (d^2 y/dt^2 )−(dy/dt)+(dy/dt)+y=0 (d^2 y/dt^2 )+y=0 y=Ae^(mt) →(d^2 y/dt^2 )=Am^2 e^(mt) Am^2 e^(mt) +Ae^(mt) =0 Ae^(mt) (m^2 +1)=0 Ae^(mt) ≠0 m^2 +1=0 m=±i y=C_1 e^(it) +C_2 e^(−it) y=C_1 (cost+isint)+C_2 (cost−isint) =(C_1 +C_2 )cost+(iC_1 −iC_2 )sint =A_1 cost+B_1 sint =A_1 cos(lnx)+B_1 sin(lnx)](https://www.tinkutara.com/question/Q60927.png)
$${x}={e}^{{t}} \rightarrow\frac{{dx}}{{dt}}={e}^{{t}} \:\:\:\:\:\:\left[{t}={lnx}\right] \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{dt}}×\frac{{dt}}{{dx}}=\frac{\mathrm{1}}{{e}^{{t}} }×\frac{{dy}}{{dt}} \\ $$$$\frac{{dy}}{{dt}}={e}^{{t}} ×\frac{{dy}}{{dx}}\rightarrow{x}\frac{{dy}}{{dx}} \\ $$$$\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right) \\ $$$$=\frac{{d}}{{dt}}\left(\frac{{dy}}{{dx}}\right)×\frac{{dt}}{{dx}} \\ $$$$=\frac{{d}}{{dt}}\left(\frac{\mathrm{1}}{{e}^{{t}} }×\frac{{dy}}{{dt}}\right)×\frac{\mathrm{1}}{{e}^{{t}} } \\ $$$$\frac{\mathrm{1}}{{e}^{{t}} }×\left[\frac{\mathrm{1}}{{e}^{{t}} }×\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }+{e}^{−{t}} ×−\mathrm{1}×\frac{{dy}}{{dt}}\right] \\ $$$$=\left(\frac{\mathrm{1}}{{e}^{{t}} }\right)^{\mathrm{2}} ×\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\left(\frac{\mathrm{1}}{{e}^{{t}} }\right)^{\mathrm{2}} ×\frac{{dy}}{{dt}} \\ $$$${so}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\left(\frac{\mathrm{1}}{{e}^{{t}} }\right)^{\mathrm{2}} \left[\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\frac{{dy}}{{dt}}\right] \\ $$$${x}^{\mathrm{2}} ×\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\frac{{dy}}{{dt}} \\ $$$${so}\:{eq}^{{n}} \: \\ $$$${x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{x}\frac{{dy}}{{dx}}+{y}=\mathrm{0} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\frac{{dy}}{{dt}}+\frac{{dy}}{{dt}}+{y}=\mathrm{0} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }+{y}=\mathrm{0} \\ $$$${y}={Ae}^{{mt}} \rightarrow\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }={Am}^{\mathrm{2}} {e}^{{mt}} \\ $$$$ \\ $$$${Am}^{\mathrm{2}} {e}^{{mt}} +{Ae}^{{mt}} =\mathrm{0} \\ $$$${Ae}^{{mt}} \left({m}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$${Ae}^{{mt}} \neq\mathrm{0} \\ $$$${m}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${m}=\pm{i} \\ $$$${y}={C}_{\mathrm{1}} {e}^{{it}} +{C}_{\mathrm{2}} {e}^{−{it}} \\ $$$${y}={C}_{\mathrm{1}} \left({cost}+{isint}\right)+{C}_{\mathrm{2}} \left({cost}−{isint}\right) \\ $$$$=\left({C}_{\mathrm{1}} +{C}_{\mathrm{2}} \right){cost}+\left({iC}_{\mathrm{1}} −{iC}_{\mathrm{2}} \right){sint} \\ $$$$={A}_{\mathrm{1}} {cost}+{B}_{\mathrm{1}} {sint} \\ $$$$={A}_{\mathrm{1}} {cos}\left({lnx}\right)+{B}_{\mathrm{1}} {sin}\left({lnx}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by necx1 last updated on 27/May/19
$${Mhen}….{I}\:{need}\:{to}\:{further}\:{my}\:{knowledge} \\ $$$${on}\:{this}.{Thanks}\:{for}\:{the}\:{help}\:{even}\:{though} \\ $$$${I}\:{dont}\:{fully}\:{understand}\:. \\ $$
Commented by necx1 last updated on 27/May/19
$${Mhen}….{I}\:{need}\:{to}\:{further}\:{my}\:{knowledge} \\ $$$${on}\:{this}.{Thanks}\:{for}\:{the}\:{help}\:{even}\:{though} \\ $$$${I}\:{dont}\:{fully}\:{understand}\:. \\ $$
Commented by tanmay last updated on 27/May/19
$${read}\:{diff}\:{eqn}\:{bookby}\:{Daniel}\:{and}\:{murry} \\ $$