The general solution of the differential equa | vedclass

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

Question

(D) Given the differential equation: $(x - (x+y) \log(x+y)) dx + x dy = 0$. Rearranging the terms: $x dy = ((x+y) \log(x+y) - x) dx$. $\frac{dy}{dx} =

Vedclass · Accepted Answer

$\log (x+y) = cx$

Vedclass · Answer

(D) Given the differential equation: $(x - (x+y) \log(x+y)) dx + x dy = 0$. Rearranging the terms: $x dy = ((x+y) \log(x+y) - x) dx$. $\frac{dy}{dx} = \frac{(x+y) \log(x+y) - x}{x}$. Let $v = x+y$,then $dv = dx + dy$,so $dy = dv - dx$. Substituting into the equation: $\frac{dv}{dx} - 1 = \frac{v \log v - x}{x} = \frac{v \log v}{x} - 1$. $\frac{dv}{dx} = \frac{v \log v}{x}$. Separating the variables: $\frac{dv}{v \log v} = \frac{dx}{x}$. Integrating both sides: $\int \frac{dv}{v \log v} = \int \frac{dx}{x}$. Let $u = \log v$,then $du = \frac{1}{v} dv$. $\int \frac{du}{u} = \int \frac{dx}{x} \implies \log|u| = \log|x| + \log|c|$. $u = cx \implies \log v = cx$. Substituting $v = x+y$ back: $\log(x+y) = cx$.

The general solution of the differential equation $(x-(x+y) \log (x+y)) dx + x dy = 0$ is

Similar Questions

The solution of $(1+y^2) dx - xy dy = 0$,$y(1)=0$ represents a conic. Its eccentricity is

The solution of the differential equation $(x + y - 1)dx + (2x + 2y - 3)dy = 0$ is

$A$ curve having the condition that the slope of the tangent at any point $(x, y)$ is two times the slope of the straight line joining the same point to the origin $(0, 0)$ is a/an:

General solution of the differential equation $\cos x(1+\cos y) dx - \sin y(1+\sin x) dy = 0$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module