A particle moving in a liquid is such that after $t$ seconds, its velocity is $v$ ms$^{-1}$, $v
e 0

Question

A particle moving in a liquid is such that after $t$ seconds, its velocity is $v$ ms$^{-1}$, $v
e 0$. $v$ satisfies the differential equation $v\frac{\mathrm{d}v}{\mathrm{d}t}+2{{v}^{2}}=t{{\mathrm{e}}^{-2t}}$. By using the substitution $y=v{{\mathrm{e}}^{2t}}$, show that $y\frac{\mathrm{d}y}{\mathrm{d}t}=t{{\mathrm{e}}^{2t}}$. [3] Hence find ${{v}^{2}}$ in terms of $t$ where $\frac{\mathrm{d}v}{\mathrm{d}t}=-8$ when $t=0$. [6]

Accepted Answer

(i) $y\frac{dy}{dt}=te^{2t}$; (ii) $v^2=te^{-2t}-\frac12e^{-2t}+\frac{33}{2}e^{-4t}$

2020 RVHS P2 Q2

2020 RVHS P2 Q2

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