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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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:

Similar Questions

The general solution $y(x)$ of the differential equation $\frac{d}{dy} \left( \int_x^y dt \right) = x$ is:

The solution of $\cos (x + y) \, dy = dx$ is

The solution of the differential equation $\frac{d\theta}{dt} = -k(\theta - \theta_0)$,where $k$ is a constant,is . . . . . .

If $y=y(x)$ is the solution of the differential equation $\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0$ with $y(0)=1$, then $y\left(\frac{\pi}{2}\right)$ is equal to

The general solution of the differential equation $\frac{dy}{dx} = \cos(x+y)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module