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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-




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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