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

$y+x^2=\frac{dy}{dx}$ has the solution

Consider the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$. If the value of $y$ is $1$ when $x = 1$,then the value of $x$ for which $y = 2$ is:

Find the general solution of the differential equation: $\cos ^{2} x \frac{d y}{d x}+y=\tan x$ where $0 \leq x < \frac{\pi}{2}$.

$A$ function $y = f(x)$ satisfies the differential equation $f(x) \sin 2x - \cos x + (1 + \sin^2 x) f'(x) = 0$ where $f(0) = 0$. Then the value of $f(\frac{\pi}{6})$ is equal to

If $x \, dy = y(dx + y \, dy)$,$y(1) = 1$,$y(x) > 0$,then $y(-3)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module