The solution of r dx + (x - r^2) dr = 0 is | vedclass

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

Question

(D) Given the differential equation $r dx + (x - r^2) dr = 0$.Rearranging the terms,we get $r dx = -(x - r^2) dr$.Dividing by $dr$,we have $r \frac{dx

Vedclass · Accepted Answer

$rx = \frac{r^3}{3} + c$

Vedclass · Answer

(D) Given the differential equation $r dx + (x - r^2) dr = 0$.Rearranging the terms,we get $r dx = -(x - r^2) dr$.Dividing by $dr$,we have $r \frac{dx}{dr} = r^2 - x$.Rearranging into the standard linear form $\frac{dx}{dr} + P(r)x = Q(r)$,we get $\frac{dx}{dr} + \frac{1}{r}x = r$.The Integrating Factor ($I$.$F$.) is $e^{\int \frac{1}{r} dr} = e^{\log r} = r$.The general solution is given by $x \cdot (I.F.) = \int Q(r) \cdot (I.F.) dr + c$.Substituting the values,$x \cdot r = \int r \cdot r dr + c$.Integrating the right side,$xr = \int r^2 dr + c$.Thus,$xr = \frac{r^3}{3} + c$.

The solution of $r dx + (x - r^2) dr = 0$ is

Similar Questions

The solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ $(x \neq 0)$ with $y(1) = 1$ is

If the general solution of the differential equation $\cos^2 x \frac{dy}{dx} + y = \tan x$ is $y = \tan x - 1 + Ce^{-\tan x}$ and it satisfies $y(\frac{\pi}{4}) = 1$,then $C =$

The solution of $dy = \cos x(2 - y \csc x)dx$ where $y = 2$ when $x = \frac{\pi}{2}$ is

The general solution of the differential equation $x \sin x \frac{dy}{dx} + (x \cos x + \sin x)y = \sin x$ is

If the slope of the tangent of the curve at any point is equal to $-y+e^{-x}$,then the equation of the curve passing through the origin is

Vedclass Test Series

Exam Paper Generator

Online Exam Module