If a curve y = f(x) passes through the point | vedclass

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

Question

(B) Given the differential equation: $y(1 + xy)dx = xdy$.Rearranging the terms: $ydx + xy^2 dx = xdy$.$xy^2 dx = xdy - ydx$.Dividing both sides by $y^

Vedclass · Accepted Answer

$\frac{4}{5}$

Vedclass · Answer

(B) Given the differential equation: $y(1 + xy)dx = xdy$.Rearranging the terms: $ydx + xy^2 dx = xdy$.$xy^2 dx = xdy - ydx$.Dividing both sides by $y^2$ (assuming $y 
eq 0$): $x dx = \frac{xdy - ydx}{y^2}$.We know that $d\left(\frac{x}{y}ight) = \frac{ydx - xdy}{y^2}$, so $\frac{xdy - ydx}{y^2} = -d\left(\frac{x}{y}ight)$.This gives: $x dx = -d\left(\frac{x}{y}ight)$.Integrating both sides: $\frac{x^2}{2} = -\frac{x}{y} + C$.Since the curve passes through $(1, -1)$, substitute $x=1, y=-1$: $\frac{1^2}{2} = -\frac{1}{-1} + C$.$\frac{1}{2} = 1 + C \Rightarrow C = -\frac{1}{2}$.So, $\frac{x^2}{2} = -\frac{x}{y} - \frac{1}{2} \Rightarrow x^2 + 1 = -\frac{2x}{y} \Rightarrow y = -\frac{2x}{x^2 + 1}$.Now, find $f\left(-\frac{1}{2}ight)$: $f\left(-\frac{1}{2}ight) = -\frac{2\left(-\frac{1}{2}ight)}{\left(-\frac{1}{2}ight)^2 + 1} = -\frac{-1}{\frac{1}{4} + 1} = \frac{1}{\frac{5}{4}} = \frac{4}{5}$.

If a curve $y = f(x)$ passes through the point $(1, -1)$ and satisfies the differential equation $y(1 + xy)dx = xdy$,then $f(-\frac{1}{2}) = $ . . . . .

Similar Questions

The solution of the differential equation,$2x^2y \frac{dy}{dx} = \tan(x^2y^2) - 2xy^2$ given $y(1) = \sqrt{\frac{\pi}{2}}$ is

The general solution of $\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y$ is

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

If the curve $y = f(x)$ passes through the point $(1, e)$ and satisfies the differential equation $dy = y(2 + \log_e x) dx, x > 0$, then $f(e)$ is equal to:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module