The solution of the differential equation ({x | vedclass

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

Question

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

Vedclass · Accepted Answer

$\log \left( {\frac{x}{y}} \right) = \frac{1}{x} + \frac{1}{y} + c$

Vedclass · Answer

(A) The given differential equation is $({x^2} - y{x^2})\frac{{dy}}{{dx}} + {y^2} + x{y^2} = 0$.Rearranging the terms,we get $x^2(1 - y)\frac{dy}{dx} = -y^2(1 + x)$.Separating the variables,we have $\frac{1 - y}{y^2} dy = -\frac{1 + x}{x^2} dx$.This can be written as $(\frac{1}{y^2} - \frac{1}{y}) dy = -(\frac{1}{x^2} + \frac{1}{x}) dx$.Integrating both sides,$\int (y^{-2} - y^{-1}) dy = -\int (x^{-2} + x^{-1}) dx$.$-y^{-1} - \log|y| = -(-x^{-1} + \log|x|) + C$.$-\frac{1}{y} - \log|y| = \frac{1}{x} - \log|x| + C$.Rearranging the terms,$\log|x| - \log|y| = \frac{1}{x} + \frac{1}{y} + C$.Thus,$\log \left( \frac{x}{y} \right) = \frac{1}{x} + \frac{1}{y} + C$.

The solution of the differential equation $({x^2} - y{x^2})\frac{{dy}}{{dx}} + {y^2} + x{y^2} = 0$ is

Similar Questions

Given that the slope of the tangent to a curve $y=y(x)$ at any point $(x, y)$ is $\frac{2y}{x^2}$. If the curve passes through the centre of the circle $x^2+y^2-2x-2y=0$,then its equation is

Let $y=y(x)$ be the solution of the differential equation $(1+y^2) e^{\tan x} dx + \cos^2 x(1+e^{2 \tan x}) dy = 0$,with $y(0)=1$. Then $y(\frac{\pi}{4})$ is equal to:

The general solution of the equation $({e^y} + 1)\cos x \, dx + {e^y}\sin x \, dy = 0$ is

The solution of the differential equation $y \, dx - x \, dy + x y^2 \, dx = 0$ is:

Find the general solution of the differential equation: $e^{x} \tan y \, dx + (1 - e^{x}) \sec^{2} y \, dy = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module