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 general solution of the differential equation $\frac{dy}{dx} = (9x + y + 5)^2$ is...

The general solution of the differential equation $\frac{d y}{d x}=\frac{x+y+1}{2 x+2 y+1}$ is

The general solution of the differential equation $\frac{dy}{dx} = \cot x \cdot \cot y$ is

The particular solution of the differential equation $\frac{dy}{dx} = \sec y$ with the initial condition $y(0) = 0$ is:

The solution of the differential equation $(1 + x^2)\frac{dy}{dx} = x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module