Let x=x(t) and y=y(t) be solutions of dxdt+ax=0 and dydt+by=0, a,bR.

Question

Let $x=x(t)$ and $y=y(t)$ be solutions of $\dfrac{dx}{dt}+ax=0$ and $\dfrac{dy}{dt}+by=0$, $a,b\in\mathbb{R}$. Given $x(0)=2$, $y(0)=1$, and $3y(1)=2x(1)$, the value of $t$ for which $x(t)=y(t)$ is:

Accepted Answer

$x(t)=2e^{-at}$, $y(t)=e^{-bt}$. From $3y(1)=2x(1)$: $3e^{-b}=4e^{-a}$.

Setting $x(t)=y(t)$: $2e^{-at}=e^{-bt}$, solved together with the ratio condition gives $t=\log\left(\dfrac43\right)^2$.

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