If x frac{dy}{dx} = y(log y - log x + 1),then | vedclass

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

Question

(D) Given the differential equation: $x \frac{dy}{dx} = y(\log y - \log x + 1)$.Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} (\log(\frac{y}{x})

Vedclass · Accepted Answer

$\log \left(\frac{y}{x}\right) = cx$,where $c$ is a constant of integration.

Vedclass · Answer

(D) Given the differential equation: $x \frac{dy}{dx} = y(\log y - \log x + 1)$.Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} (\log(\frac{y}{x}) + 1)$.This is a homogeneous differential equation. Let $v = \frac{y}{x}$,so $y = vx$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = v(\log v + 1)$.$v + x \frac{dv}{dx} = v \log v + v$.$x \frac{dv}{dx} = v \log v$.Separating the variables: $\frac{dv}{v \log v} = \frac{dx}{x}$.Integrating both sides: $\int \frac{1}{v \log v} dv = \int \frac{1}{x} dx$.Let $u = \log v$,then $du = \frac{1}{v} dv$. The integral becomes $\int \frac{1}{u} du = \int \frac{1}{x} dx$.$\log(u) = \log(x) + \log(c)$,where $\log(c)$ is the constant of integration.$\log(\log v) = \log(cx)$.Taking the exponential of both sides: $\log v = cx$.Substituting $v = \frac{y}{x}$ back: $\log(\frac{y}{x}) = cx$.

If $x \frac{dy}{dx} = y(\log y - \log x + 1)$,then the general solution of this equation is

Similar Questions

Let $f(x) = \sqrt{\lim_{r \rightarrow x} \left\{ \frac{2r^2 \left[(f(r))^2 - f(x)f(r)\right]}{r^2 - x^2} - r^3 e^{\frac{f(r)}{r}} \right\}}$ be differentiable in $(-\infty, 0) \cup (0, \infty)$ and $f(1) = 1$. Then the value of $ea$,such that $f(a) = 0$,is equal to:

Let $y(x)$ be the solution of the differential equation $x^2 \frac{dy}{dx} + xy = x^2 + y^2$,$x > \frac{1}{e}$,satisfying $y(1) = 0$. Then the value of $2 \frac{(y(e))^2}{y(e^2)}$ is $....$

The solution of the differential equation $x + y\frac{dy}{dx} = 2y$ is

If $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$ and $y(1) = 1$,then find the value of $x$ that satisfies $y(x) = e$.

$(x^2 + y^2)dy = xy dx$. If $y(x_0) = e$ and $y(1) = 1$,then the value of $x_0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module