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Question Number 40920 by ajfour last updated on 29/Jul/18

(d^( 2) x/dt^2 )=a−b(1−(l/(√(x^2 +l^2 ))))x Find x(t) if x(0)=x_0 , x′(0)=0 .

$$\frac{{d}^{\:\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }={a}−{b}\left(\mathrm{1}−\frac{{l}}{\sqrt{{x}^{\mathrm{2}} +{l}^{\mathrm{2}} }}\right){x}\: \\ $$$${Find}\:{x}\left({t}\right)\:{if}\:{x}\left(\mathrm{0}\right)={x}_{\mathrm{0}} \:,\:{x}'\left(\mathrm{0}\right)=\mathrm{0}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jul/18

v(dv/dx)=a−b(1−(l/(√(x^2 +l^2 ))))x vdv=(a−bx+((blx)/(√(x^2 +l^2 ))) ) dx vdv=(a−bx)dx+((bl)/2)((d(x^2 +l^2 ))/(√(x^2 +l^2 ))) intregating (v^2 /2)=ax−b(x^2 /2)+((bl)/2)(((√(x^2 +l^2 )) )/(1/2))+c v=(dx/dt)=x′ (v^2 /2)=ax−((bx^2 )/2)+bl(√(x^2 +l^2 )) +c boundary condition at t=0 (dx/dt)=v=0 x=x_0 0=ax_0 −((bx_0 ^2 )/2)+bl(√(x_0 ^2 +l^2 )) +c (v^2 /2)=a(x−x_0 )−(b^2 /2)(x^2 −x_0 ^2 )+bl((√(x^2 +l^2 )) −(√(x_0 ^2 +l^2 )) contd...

$${v}\frac{{dv}}{{dx}}={a}−{b}\left(\mathrm{1}−\frac{{l}}{\sqrt{{x}^{\mathrm{2}} +{l}^{\mathrm{2}} }}\right){x} \\ $$$${vdv}=\left({a}−{bx}+\frac{{blx}}{\sqrt{{x}^{\mathrm{2}} +{l}^{\mathrm{2}} }}\:\:\right)\:{dx} \\ $$$${vdv}=\left({a}−{bx}\right){dx}+\frac{{bl}}{\mathrm{2}}\frac{{d}\left({x}^{\mathrm{2}} +{l}^{\mathrm{2}} \right)}{\sqrt{{x}^{\mathrm{2}} +{l}^{\mathrm{2}} }} \\ $$$${intregating} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}={ax}−{b}\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{bl}}{\mathrm{2}}\frac{\sqrt{{x}^{\mathrm{2}} +{l}^{\mathrm{2}} }\:}{\frac{\mathrm{1}}{\mathrm{2}}}+{c} \\ $$$${v}=\frac{{dx}}{{dt}}={x}'\: \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}={ax}−\frac{{bx}^{\mathrm{2}} }{\mathrm{2}}+{bl}\sqrt{{x}^{\mathrm{2}} +{l}^{\mathrm{2}} }\:\:+{c} \\ $$$${boundary}\:{condition} \\ $$$${at}\:{t}=\mathrm{0}\:\:\frac{{dx}}{{dt}}={v}=\mathrm{0}\:\:\:{x}={x}_{\mathrm{0}} \\ $$$$\mathrm{0}={ax}_{\mathrm{0}} −\frac{{bx}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}}+{bl}\sqrt{{x}_{\mathrm{0}} ^{\mathrm{2}} +{l}^{\mathrm{2}} }\:+{c} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}={a}\left({x}−{x}_{\mathrm{0}} \right)−\frac{{b}^{\mathrm{2}} }{\mathrm{2}}\left({x}^{\mathrm{2}} −{x}_{\mathrm{0}} ^{\mathrm{2}} \right)+{bl}\left(\sqrt{{x}^{\mathrm{2}} +{l}^{\mathrm{2}} }\:\:−\sqrt{{x}_{\mathrm{0}} ^{\mathrm{2}} +{l}^{\mathrm{2}} }\right. \\ $$$$ \\ $$$${contd}... \\ $$

Commented by ajfour last updated on 29/Jul/18

thanks sir for as far ..

$${thanks}\:{sir}\:{for}\:{as}\:{far}\:.. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Jul/18

(√(2{(((−b)/2)x^2 +ax+(b^2 /2)x_0 ^(2 ) −ax_0 −bl(√(x_0 ^2 +l^2 )) +bl(√(x^2 +l^2 )) )) the whole expression =(dx/dt) (dx/dt)=(√(2(Ax^2 +Bx+C+D(√(x^2 +l^2 )) )) form x=∫(√(2(Ax^2 +Bx+C+D(√(x^2 +l^2 )) )) dt

$$\sqrt{\mathrm{2}\left\{\left(\frac{−{b}}{\mathrm{2}}{x}^{\mathrm{2}} +{ax}+\frac{{b}^{\mathrm{2}} }{\mathrm{2}}{x}_{\mathrm{0}} ^{\mathrm{2}\:} −{ax}_{\mathrm{0}} −{bl}\sqrt{{x}_{\mathrm{0}} ^{\mathrm{2}} +{l}^{\mathrm{2}} }\:+{bl}\sqrt{{x}^{\mathrm{2}} +{l}^{\mathrm{2}} }\:\:\:\right.\right.} \\ $$$${the}\:{whole}\:{expression}\:=\frac{{dx}}{{dt}} \\ $$$$\frac{{dx}}{{dt}}=\sqrt{\mathrm{2}\left({Ax}^{\mathrm{2}} +{Bx}+{C}+{D}\sqrt{{x}^{\mathrm{2}} +{l}^{\mathrm{2}} }\:\right.}\:\:\:{form} \\ $$$${x}=\int\sqrt{\mathrm{2}\left({Ax}^{\mathrm{2}} +{Bx}+{C}+{D}\sqrt{{x}^{\mathrm{2}} +{l}^{\mathrm{2}} }\:\right.}\:\:{dt} \\ $$$$\:\:\: \\ $$$$\:\:\:\: \\ $$

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