Let y=y(t) be a solution of the differential | vedclass

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

Question

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dt} + P(t)y = Q(t)$,where $P(t) = \alpha$ and $Q(t) = \ga

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dt} + P(t)y = Q(t)$,where $P(t) = \alpha$ and $Q(t) = \gamma e^{-\beta t}$.The integrating factor ($I$.$F$.) is given by $e^{\int P(t) dt} = e^{\int \alpha dt} = e^{\alpha t}$.The general solution is $y \cdot (I.F.) = \int Q(t) \cdot (I.F.) dt + C$.Substituting the values,we get $y e^{\alpha t} = \int \gamma e^{-\beta t} \cdot e^{\alpha t} dt + C = \gamma \int e^{(\alpha - \beta)t} dt + C$.Case $1$: If $\alpha 
eq \beta$,then $y e^{\alpha t} = \frac{\gamma}{\alpha - \beta} e^{(\alpha - \beta)t} + C$,which implies $y(t) = \frac{\gamma}{\alpha - \beta} e^{-\beta t} + C e^{-\alpha t}$.Since $\alpha > 0$ and $\beta > 0$,as $t ightarrow \infty$,$e^{-\beta t} ightarrow 0$ and $e^{-\alpha t} ightarrow 0$.Therefore,$\lim_{t ightarrow \infty} y(t) = 0 + 0 = 0$.Case $2$: If $\alpha = \beta$,then $y e^{\alpha t} = \int \gamma dt + C = \gamma t + C$,which implies $y(t) = \gamma t e^{-\alpha t} + C e^{-\alpha t}$.Using $L$'Hopital's rule,$\lim_{t ightarrow \infty} \frac{\gamma t}{e^{\alpha t}} = \lim_{t ightarrow \infty} \frac{\gamma}{\alpha e^{\alpha t}} = 0$.Thus,in both cases,$\lim_{t ightarrow \infty} y(t) = 0$.

Let $y=y(t)$ be a solution of the differential equation $\frac{dy}{dt}+\alpha y=\gamma e^{-\beta t}$,where $\alpha > 0, \beta > 0$ and $\gamma > 0$. Then $\lim_{t \rightarrow \infty} y(t)$ is:

Similar Questions

$A$ function $y=f(x)$ satisfies $f(x) \sin 2x + \sin x - (1 + \cos^2 x) f'(x) = 0$ with the condition $f(0) = 0$. Then $f(\frac{\pi}{2})$ is equal to

If a curve passes through the origin and the slope of the tangent to it at any point $(x, y)$ is $\frac{x^{2}-4x+y+8}{x-2}$,then this curve also passes through the point

Let $y=y(x)$ be the solution of the differential equation $x\frac{dy}{dx}-\sin(2y)=x^{3}(2-x^{3})\cos^{2}y,$ for $x\ne0.$ If $y(2)=0,$ then $\tan(y(1))$ is equal to

Find the general solution of the differential equation $\frac{dy}{dx}-y=\cos x$.

Let $y=y(x)$ be the solution of the differential equation $(x \log x) \frac{dy}{dx} + y = 2x \log x$ for $x \geq 1$. Then $y(e)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module