Question

The differential equation of a free falling body is governed by the DE $v\frac{dv}{dx}=-cx-b{v}^2$.where $v$ and $x$ are velocity and displacement respectively. The velocity of the body is given by the equation $v^2=\frac{c}{2b^2}(1-e^{-2bx})-\frac{cx}{b}$

• A. True
• B. False

Answer 1

The correct option is A True

We have $v\frac{dv}{dx}=-cx-b{v}^2$

Putting $v^2=y,v\frac{dv}{dx}=\frac{1}{2}\frac{dy}{dx}$

$\therefore \frac{1}{2}.\frac{dy}{dx}+by=-cx$ $\therefore \frac{dy}{dx}+2by=-2cx$

This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$

$\therefore I.F=e^{\int Pdx}=e^{2bdx}=e^{2bx}$

$\therefore$ The solution is $y{e}^{2bx}=\int e^{2bx}(-2cx)dx+c^{\prime }$

$y{e}^{2bx}=-2c[x\cdot \frac{e^{2bx}}{2b}-\int \frac{e^{2bx}}{2b}\cdot (1)\cdot dx]+c^{\prime }$ (integrate by parts)

$=-2c[x\cdot \frac{e^{2bx}}{2b}-\frac{e^{2bx}}{4b^2}]+c^{\prime }$

Resubstituting $y=v^2$,

$v^2{e}^{2bx}=-\frac{cx}{b}\cdot e^{2bx}+\frac{c}{2b^2}{e}^{2bx}+c^{\prime }$ ...(1)

By data, when $x=0,v=0$ $\therefore c^{\prime }=-\frac{c}{2b^2}$

Therefore, $v^2{e}^{2bx}=-\frac{cx}{b}{e}^{2bx}+\frac{c}{2b^2}{e}^{2bx}-\frac{c}{2b^2}$

$\Rightarrow v^2=-\frac{cx}{b}+\frac{c}{2b^2}-\frac{c}{2b^2}{e}^{-2bx}$

$=\frac{c}{2b^2}(1-e^{-2bx})-\frac{cx}{b}$.