If x frac{dy}{dx} + y = x frac{f(xy)}{f'(xy)} | vedclass

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

Question

(A) Given the differential equation: $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$.We know that $x dy + y dx = d(xy)$.Rearranging the given equation:

Vedclass · Accepted Answer

$k e^{x^2/2}$

Vedclass · Answer

(A) Given the differential equation: $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$.We know that $x dy + y dx = d(xy)$.Rearranging the given equation: $x dy + y dx = x \frac{f(xy)}{f'(xy)} dx$.This is not quite right,let's rewrite: $\frac{dy}{dx} + \frac{y}{x} = \frac{f(xy)}{f'(xy)}$.Multiply by $x$: $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$.Note that $x \frac{dy}{dx} + y = \frac{d}{dx}(xy)$.So,$\frac{d(xy)}{dx} = x \frac{f(xy)}{f'(xy)}$.Rearranging the terms: $\frac{f'(xy)}{f(xy)} d(xy) = x dx$.Integrating both sides: $\int \frac{f'(xy)}{f(xy)} d(xy) = \int x dx$.$\ln|f(xy)| = \frac{x^2}{2} + C$.Taking the exponential of both sides: $f(xy) = e^{\frac{x^2}{2} + C} = k e^{x^2/2}$,where $k = e^C$.

If $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$,then $f(xy)$ is equal to

Similar Questions

The solution of the differential equation $(2x - 3y + 5)dx + (9y - 6x - 7)dy = 0$ is

The solution of the differential equation $ydx - xdy = x^2 ydx$ is

The general solution of $\frac{dy}{dx} = \cos^2(x-y-1)$ is given by $x=$

$A$ particle starts at the origin and moves along the $x$-axis in such a way that its velocity at the point $(x, 0)$ is given by the formula $\frac{dx}{dt} = \cos^2(\pi x)$. Then the particle never reaches the point on:

The solution of the differential equation $\frac{dy}{dx} = \frac{y - x}{y - x - 1}$,given $y(-5) = -5$,represents:

Vedclass Test Series

Exam Paper Generator

Online Exam Module