Let x=x(t) and y=y(t) be solutions of the dif | vedclass

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(D) The given differential equations are $\frac{dx}{dt} = -ax$ and $\frac{dy}{dt} = -by$.Solving $\frac{dx}{dt} = -ax$ by separating variables,we get

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$\log_{\frac{4}{3}} 2$

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(D) The given differential equations are $\frac{dx}{dt} = -ax$ and $\frac{dy}{dt} = -by$.Solving $\frac{dx}{dt} = -ax$ by separating variables,we get $\int \frac{dx}{x} = -\int a dt$,which gives $\ln|x| = -at + C_1$.Using $x(0)=2$,we find $\ln 2 = C_1$,so $x(t) = 2e^{-at}$.Similarly,solving $\frac{dy}{dt} = -by$ with $y(0)=1$,we get $y(t) = e^{-bt}$.Given $3y(1) = 2x(1)$,we substitute the expressions:$3e^{-b} = 2(2e^{-a}) \implies 3e^{-b} = 4e^{-a} \implies e^{a-b} = \frac{4}{3}$.We want to find $t$ such that $x(t) = y(t)$:$2e^{-at} = e^{-bt} \implies 2 = e^{(a-b)t}$.Taking the natural logarithm on both sides:$\ln 2 = (a-b)t$.Since $e^{a-b} = \frac{4}{3}$,we have $a-b = \ln(\frac{4}{3})$.Thus,$\ln 2 = t \ln(\frac{4}{3}) \implies t = \frac{\ln 2}{\ln(\frac{4}{3})} = \log_{\frac{4}{3}} 2$.

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

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