Find the general 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: $y \log y \, dx - x \, dy = 0$. Rearranging the terms,we get: $y \log y \, dx = x \, dy$. Separating the varia

Vedclass · Accepted Answer

$y = e^{Cx}$

Vedclass · Answer

(A) The given differential equation is: $y \log y \, dx - x \, dy = 0$. Rearranging the terms,we get: $y \log y \, dx = x \, dy$. Separating the variables,we have: $\frac{dy}{y \log y} = \frac{dx}{x}$. Integrating both sides: $\int \frac{dy}{y \log y} = \int \frac{dx}{x}$. Let $t = \log y$,then $dt = \frac{1}{y} \, dy$. Substituting this into the integral: $\int \frac{dt}{t} = \int \frac{dx}{x}$. Integrating,we get: $\log |t| = \log |x| + \log |C|$. $\log |\log y| = \log |Cx|$. Taking the exponential of both sides: $\log y = Cx$. Thus,the general solution is: $y = e^{Cx}$.

Find the general solution of the differential equation: $y \log y \, dx - x \, dy = 0$.

Similar Questions

The solution of $\frac{dy}{dx} + 1 = e^{x+y}$ is

If $f(x)$ is a function such that $f^{\prime}(x)=\sqrt{f^2(x)-1}$ and $f(0)=1$,then $f(1)=$

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

If $y = y(x)$ is the solution of the differential equation $2x^{2} \frac{dy}{dx} - 2xy + 3y^{2} = 0$ such that $y(e) = \frac{e}{3}$,then $y(1)$ is equal to

General solution of the differential equation $x \cos y \,dy = (x e^x \log x + e^x) dx$ is (where $C$ is a constant of integration.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module