Consider the differential equation y^2 dx + ( | vedclass

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

Question

(C) The given differential equation is $y^2 dx + (x - \frac{1}{y}) dy = 0$.Rearranging the terms,we get $y^2 dx + x dy = \frac{1}{y} dy$.Dividing by $

Vedclass · Accepted Answer

$\frac{3}{2} - \frac{1}{\sqrt{e}}$

Vedclass · Answer

(C) The given differential equation is $y^2 dx + (x - \frac{1}{y}) dy = 0$.Rearranging the terms,we get $y^2 dx + x dy = \frac{1}{y} dy$.Dividing by $y^2 dy$,we obtain $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{y^2}$ and $Q(y) = \frac{1}{y^3}$.The integrating factor $(IF)$ is $e^{\int P(y) dy} = e^{\int \frac{1}{y^2} dy} = e^{-\frac{1}{y}}$.The general solution is $x \cdot e^{-\frac{1}{y}} = \int \frac{1}{y^3} e^{-\frac{1}{y}} dy + C$.Let $t = -\frac{1}{y}$,then $dt = \frac{1}{y^2} dy$. Also,$\frac{1}{y} = -t$.Substituting these,$\int (-t) e^t dt = - (t e^t - e^t) = e^t(1 - t) = e^{-\frac{1}{y}}(1 + \frac{1}{y})$.Thus,$x e^{-\frac{1}{y}} = e^{-\frac{1}{y}}(1 + \frac{1}{y}) + C$.Given $y=1$ when $x=1$,we have $1 \cdot e^{-1} = e^{-1}(1 + 1) + C \implies e^{-1} = 2e^{-1} + C \implies C = -e^{-1}$.So,$x e^{-\frac{1}{y}} = e^{-\frac{1}{y}}(1 + \frac{1}{y}) - e^{-1}$.For $y=2$,$x e^{-1/2} = e^{-1/2}(1 + 1/2) - e^{-1} = \frac{3}{2} e^{-1/2} - e^{-1}$.Dividing by $e^{-1/2}$,we get $x = \frac{3}{2} - e^{-1/2} = \frac{3}{2} - \frac{1}{\sqrt{e}}$.

Consider the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$. If the value of $y$ is $1$ when $x = 1$,then the value of $x$ for which $y = 2$ is:

Similar Questions

The integrating factor of the differential equation $\sin x \frac{dy}{dx} - y \cos x = 1$ is

Find the general solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ where $x \neq 0$.

If $\frac{dy}{dx} + \frac{y}{x} = x^2$,then $2y(2) - y(1) =$

Let $f$ be a differentiable function with $\lim _{x \rightarrow \infty} f(x)=0$. If $y^{\prime}+y f^{\prime}(x)-f(x) f^{\prime}(x)=0$ and $\lim _{x \rightarrow \infty} y(x)=0$, then:

If $y=y(x)$ is the solution curve of the differential equation $\frac{dy}{dx} + y \tan x = x \sec x$,$0 \leq x \leq \frac{\pi}{3}$,with $y(0)=1$,then $y\left(\frac{\pi}{6}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module