The integrating factor of x frac{dy}{dx} - 2y | vedclass

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

Question

(B) The given differential equation is $x \frac{dy}{dx} - 2y = x^2 + \sin \left( \frac{1}{x^2} \right)$.Divide the entire equation by $x$ to write it

Vedclass · Accepted Answer

$\frac{1}{x^2}$

Vedclass · Answer

(B) The given differential equation is $x \frac{dy}{dx} - 2y = x^2 + \sin \left( \frac{1}{x^2} \right)$.Divide the entire equation by $x$ to write it in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:$\frac{dy}{dx} - \frac{2}{x}y = x + \frac{1}{x} \sin \left( \frac{1}{x^2} \right)$.Here,$P(x) = -\frac{2}{x}$.The integrating factor $(IF)$ is given by the formula $IF = e^{\int P(x) dx}$.$IF = e^{\int -\frac{2}{x} dx} = e^{-2 \ln|x|} = e^{\ln|x^{-2}|} = x^{-2} = \frac{1}{x^2}$.

The integrating factor of $x \frac{dy}{dx} - 2y = x^2 + \sin \left( \frac{1}{x^2} \right)$ is

Similar Questions

The solution of $(y-3 x^2) d x+x d y=0$ is

If the solution $y(x)$ of the differential equation $\sin x \frac{dy}{dx} + y \cos x = e^{2x}, x \in (0, \pi)$ satisfies $y\left(\frac{\pi}{2}\right) = 0$,then $y\left(\frac{\pi}{6}\right) = $

Let $x = x(y)$ be the solution of the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$. If $x(1) = 1$,then $x(\frac{1}{2})$ is:

Find the general solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ where $x \neq 0$.

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{5}{x(x^5+1)}y=\frac{(x^5+1)^2}{x^7}$,for $x > 0$. If $y(1)=2$,then $y(2)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module