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Question Number 43616 by Tawa1 last updated on 12/Sep/18

Solve:   x^2 y′′  +  xy′  − y = x ,        y_1  = x

$$\mathrm{Solve}:\:\:\:\mathrm{x}^{\mathrm{2}} \mathrm{y}''\:\:+\:\:\mathrm{xy}'\:\:−\:\mathrm{y}\:=\:\mathrm{x}\:,\:\:\:\:\:\:\:\:\mathrm{y}_{\mathrm{1}} \:=\:\mathrm{x} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Sep/18

x=e^t    (dx/dt)=e^t   (dy/dx)=(dy/dt)×(dt/dx)=(dy/dt)×(1/e^t )  (dy/dt)=e^t .(dy/dx)=x(dy/dx)  (d^2 y/dx^2 )=(d/dx)((dy/dx))=(d/dt)((1/e^t )(dy/dt))×(dt/dx)  =((1/e^t )×(d^2 y/dt^2 )−e^(−t) ×(dy/dt))×(dt/dx)  =(1/e^t )((d^2 y/dt^2 )−(dy/dt))×(1/e^t )  (e^t )^2 ×(d^2 y/dx^2 )=((d^2 y/dt^2 )−(dy/dt))  x^2 ×(d^2 y/dx^2 )=(d^2 y/dt^2 )−(dy/dt)  now substituting the value obtained  (d^2 y/dt^2 )−(dy/dt)+(dy/dt)−y=e^t   (d^2 y/dt^2 )−y=e^t   let   y=e^(mt)  is a solution  m^2 e^(mt) −e^(mt) =0  e^(mt) (m^2 −1)=0  e^(my)  can not be zero  m=±1  c.F=Ae^t +Be^(−t)   P.I=(e^t /(D^2 −1))  =(e^t /((D+1)(D−1)))  =(1/((1+1)))×(e^t /)×(1/(D+1−1)).1  =(e^t /2)×t  complete solution  =Ae^t +B^ e^(−t) +((te^t )/2)  =Ax+Bx^(−1) +((x×lnx)/2)

$${x}={e}^{{t}} \:\:\:\frac{{dx}}{{dt}}={e}^{{t}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{dt}}×\frac{{dt}}{{dx}}=\frac{{dy}}{{dt}}×\frac{\mathrm{1}}{{e}^{{t}} } \\ $$$$\frac{{dy}}{{dt}}={e}^{{t}} .\frac{{dy}}{{dx}}={x}\frac{{dy}}{{dx}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right)=\frac{{d}}{{dt}}\left(\frac{\mathrm{1}}{{e}^{{t}} }\frac{{dy}}{{dt}}\right)×\frac{{dt}}{{dx}} \\ $$$$=\left(\frac{\mathrm{1}}{{e}^{{t}} }×\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−{e}^{−{t}} ×\frac{{dy}}{{dt}}\right)×\frac{{dt}}{{dx}} \\ $$$$=\frac{\mathrm{1}}{{e}^{{t}} }\left(\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\frac{{dy}}{{dt}}\right)×\frac{\mathrm{1}}{{e}^{{t}} } \\ $$$$\left({e}^{{t}} \right)^{\mathrm{2}} ×\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\left(\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\frac{{dy}}{{dt}}\right) \\ $$$${x}^{\mathrm{2}} ×\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\frac{{dy}}{{dt}} \\ $$$${now}\:{substituting}\:{the}\:{value}\:{obtained} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\frac{{dy}}{{dt}}+\frac{{dy}}{{dt}}−{y}={e}^{{t}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−{y}={e}^{{t}} \\ $$$${let}\:\:\:{y}={e}^{{mt}} \:{is}\:{a}\:{solution} \\ $$$${m}^{\mathrm{2}} {e}^{{mt}} −{e}^{{mt}} =\mathrm{0} \\ $$$${e}^{{mt}} \left({m}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$${e}^{{my}} \:{can}\:{not}\:{be}\:{zero} \\ $$$${m}=\pm\mathrm{1} \\ $$$${c}.{F}={Ae}^{{t}} +{Be}^{−{t}} \\ $$$${P}.{I}=\frac{{e}^{{t}} }{{D}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\frac{{e}^{{t}} }{\left({D}+\mathrm{1}\right)\left({D}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{1}\right)}×\frac{{e}^{{t}} }{}×\frac{\mathrm{1}}{{D}+\mathrm{1}−\mathrm{1}}.\mathrm{1} \\ $$$$=\frac{{e}^{{t}} }{\mathrm{2}}×{t} \\ $$$${complete}\:{solution} \\ $$$$={Ae}^{{t}} +{B}^{} {e}^{−{t}} +\frac{{te}^{{t}} }{\mathrm{2}} \\ $$$$={Ax}+{Bx}^{−\mathrm{1}} +\frac{{x}×{lnx}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

Commented by Tawa1 last updated on 13/Sep/18

God bless you sir

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

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