If frac{dy}{dx} = frac{xy + y}{xy + x},then t | vedclass

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

Question

(C) Given the differential equation: $\frac{dy}{dx} = \frac{y(x + 1)}{x(y + 1)}$Separating the variables,we get:$\frac{y + 1}{y} dy = \frac{x + 1}{x}

Vedclass · Accepted Answer

$y = Ax{e^{x - y}}$

Vedclass · Answer

(C) Given the differential equation: $\frac{dy}{dx} = \frac{y(x + 1)}{x(y + 1)}$Separating the variables,we get:$\frac{y + 1}{y} dy = \frac{x + 1}{x} dx$This can be written as:$(1 + \frac{1}{y}) dy = (1 + \frac{1}{x}) dx$Integrating both sides:$\int (1 + \frac{1}{y}) dy = \int (1 + \frac{1}{x}) dx$$y + \log|y| = x + \log|x| + C$Rearranging the terms:$y - x = \log|x| - \log|y| + C$$y - x = \log|\frac{x}{y}| + C$Let $C = -\log|A|$,then:$y - x = \log|\frac{x}{y}| - \log|A| = \log|\frac{x}{Ay}|$Taking the exponential of both sides:$e^{y - x} = \frac{x}{Ay}$$Ay e^{y - x} = x$$y = \frac{x}{A} e^{x - y}$Replacing $\frac{1}{A}$ with a new constant $A$,we get:$y = Ax e^{x - y}$

If $\frac{dy}{dx} = \frac{xy + y}{xy + x}$,then the solution of the differential equation is

Similar Questions

Along the $X$-axis,three charges $\frac{q}{2}, -q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2a$ respectively. The resultant electric potential at $x=a+r$ (if $a \ll r$) is: ($\varepsilon_0$ is the permittivity of free space)

$A$ common emitter transistor amplifier has a current gain of $50$. If the load resistance is $4 \ k\Omega$ and input resistance is $500 \ \Omega$,the voltage gain of the amplifier is

Nuclear fusion is common to the pair:

Which one of the following is a correct set?

$A$ lift is moving downwards with an acceleration equal to the acceleration due to gravity. $A$ body of mass $M$ kept on the floor of the lift is pulled horizontally. If the coefficient of friction is $\mu$,then the frictional resistance offered by the body is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module