Let x = x(y) be the solution of the different | vedclass

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

Question

(C) The given differential equation is $y^2 dx + (x - \frac{1}{y}) dy = 0$. Dividing by $y^2 dy$,we get $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}

Vedclass · Accepted Answer

$3 - e$

Vedclass · Answer

(C) The given differential equation is $y^2 dx + (x - \frac{1}{y}) dy = 0$. Dividing by $y^2 dy$,we get $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$. 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 ($I$.$F$.) is $e^{\int P(y) dy} = e^{\int y^{-2} dy} = e^{-1/y}$. The solution is $x \cdot e^{-1/y} = \int Q(y) \cdot e^{-1/y} dy + C$. Let $t = -1/y$,then $dt = \frac{1}{y^2} dy$. $x \cdot e^{-1/y} = \int (-t) e^t dt + C = -(t e^t - e^t) + C = e^t(1 - t) + C$. Substituting $t = -1/y$,we get $x \cdot e^{-1/y} = e^{-1/y}(1 + \frac{1}{y}) + C$. Given $x(1) = 1$,we have $1 \cdot e^{-1} = e^{-1}(1 + 1) + C$,so $e^{-1} = 2e^{-1} + C$,which implies $C = -e^{-1}$. Thus,$x = 1 + \frac{1}{y} - e^{1/y} \cdot e^{-1} = 1 + \frac{1}{y} - e^{(1/y) - 1}$. For $y = 1/2$,$x = 1 + \frac{1}{1/2} - e^{(1/(1/2)) - 1} = 1 + 2 - e^{2-1} = 3 - e$.

Let $x = x(y)$ be the solution of the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$. If $x(1) = 1$,then $x(\frac{1}{2})$ is:

Similar Questions

Find the solution of the differential equation given below:
$\frac{dy}{dx} + y \cdot \csc^2 (x) = \csc^2 (x) \cdot \cot (x)$

If $\cos x \frac{dy}{dx} - y \sin x = 6x$,where $0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{3}) = 0$,then find $y(\frac{\pi}{6})$.

The solution of the differential equation $\frac{dy}{dx} + 2y \cot x = 3x^2 \csc^2 x$ is

An integrating factor of the differential equation $\left(1-x^2\right) \frac{d y}{d x}+x y=\frac{x^4}{\left(1+x^5\right)}\left(\sqrt{1-x^2}\right)^3$ is

Solve the following differential equation: $\left(x^2+1\right) \frac{dy}{dx} + 4xy = \frac{1}{x^2+1}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module