The homogeneous differential equation of the | vedclass

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

Question

(C) The given differential equation is $\left(1+e^{\frac{x}{y}}\right) dx + e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) dy = 0$. Rearranging the terms,w

Vedclass · Accepted Answer

$x=vy$

Vedclass · Answer

(C) The given differential equation is $\left(1+e^{\frac{x}{y}}\right) dx + e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) dy = 0$. Rearranging the terms,we get $\frac{dx}{dy} = -\frac{e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}}$. Since the equation involves the term $\frac{x}{y}$,it is a homogeneous differential equation of the form $\frac{dx}{dy} = F\left(\frac{x}{y}\right)$. To solve such an equation,we substitute $x = vy$,where $v$ is a function of $y$. Therefore,the correct substitution is $x = vy$.

The homogeneous differential equation of the form $\left(1+e^{\frac{x}{y}}\right) dx + e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) dy = 0$ can be solved by making the substitution:

Similar Questions

If $y' = \frac{x - y}{x + y}$,then its solution is

In a single throw of two dice,what is the probability of obtaining a sum of numbers greater than $7$,given that $4$ appears on the first die?

Show that the differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$ is homogeneous and solve it.

If $\frac{dy}{dx} = f(x, y)$ is a homogeneous differential equation,then the general form of $f(x, y)$ is

Which one of the following is a homogeneous differential equation?

Vedclass Test Series

Exam Paper Generator

Online Exam Module