Solve xdx + ydy = frac{xdy - ydx}{x^2 + y^2} | vedclass

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

Question

(A) The given differential equation is $xdx + ydy = \frac{xdy - ydx}{x^2 + y^2}$.We know that $d(x^2 + y^2) = 2xdx + 2ydy$,which implies $xdx + ydy =

Vedclass · Accepted Answer

$\frac{1}{2}(x^2 + y^2) = 	an^{-1}(y/x) + c$

Vedclass · Answer

(A) The given differential equation is $xdx + ydy = \frac{xdy - ydx}{x^2 + y^2}$.We know that $d(x^2 + y^2) = 2xdx + 2ydy$,which implies $xdx + ydy = \frac{1}{2} d(x^2 + y^2)$.Also,we know that $d(	an^{-1}(y/x)) = \frac{1}{1 + (y/x)^2} \cdot d(y/x) = \frac{x^2}{x^2 + y^2} \cdot \frac{xdy - ydx}{x^2} = \frac{xdy - ydx}{x^2 + y^2}$.Substituting these into the given equation,we get:$\frac{1}{2} d(x^2 + y^2) = d(	an^{-1}(y/x))$.Integrating both sides,we get:$\frac{1}{2}(x^2 + y^2) = 	an^{-1}(y/x) + c$.

Solve $xdx + ydy = \frac{xdy - ydx}{x^2 + y^2}$

Similar Questions

The solution of the differential equation $2 x \left(\frac{d y}{d x}\right) - y = 4$ represents a family of

Let a smooth curve $y=f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\left(\frac{-y}{x}\right)$. If the curve passes through the points $(1, 2)$ and $(8, 1)$,then $\left| y \left(\frac{1}{8}\right) \right|$ is equal to

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 $\frac{dy}{dx} + \sin \left(\frac{x+y}{2}\right) = \sin \left(\frac{x-y}{2}\right)$ is

The solution of the differential equation $y \, dx - x \, dy + x y^2 \, dx = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module