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

$$\frac{d^{2}x}{{dt}^2}=a-b(1-\frac{l}{\sqrt{x^2+l^2}})x$$$$Findx\left(t\right)ifx\left(0\right)=x_0,x^{\prime }\left(0\right)=0.$$

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

$$v\frac{dv}{dx}=a-b(1-\frac{l}{\sqrt{x^2+l^2}})x$$$$vdv=(a-bx+\frac{blx}{\sqrt{x^2+l^2}})dx$$$$vdv=(a-bx)dx+\frac{bl}{2}\frac{d(x^2+l^2)}{\sqrt{x^2+l^2}}$$$$intregating$$$$\frac{v^2}{2}=ax-b\frac{x^2}{2}+\frac{bl}{2}\frac{\sqrt{x^2+l^2}}{\frac{1}{2}}+c$$$$v=\frac{dx}{dt}=x^{\prime }$$$$\frac{v^2}{2}=ax-\frac{{bx}^2}{2}+bl\sqrt{x^2+l^2}+c$$$$boundarycondition$$$$att=0\frac{dx}{dt}=v=0x=x_0$$$$0={ax}_0-\frac{{bx}_0^2}{2}+bl\sqrt{x_0^2+l^2}+c$$$$\frac{v^2}{2}=a(x-x_0)-\frac{b^2}{2}(x^2-x_0^2)+bl(\sqrt{x^2+l^2}-\sqrt{x_0^2+l^2}$$$$$$$$contd...$$

Commented by ajfour last updated on 29/Jul/18

$$thankssirforasfar..$$

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

$$\sqrt{2\{(\frac{-b}{2}{x}^2+ax+\frac{b^2}{2}{x}_0^2-{ax}_0-bl\sqrt{x_0^2+l^2}+bl\sqrt{x^2+l^2}}$$$$thewholeexpression=\frac{dx}{dt}$$$$\frac{dx}{dt}=\sqrt{2({Ax}^2+Bx+C+D\sqrt{x^2+l^2}}form$$$$x=\int \sqrt{2({Ax}^2+Bx+C+D\sqrt{x^2+l^2}}dt$$$$$$$$$$

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