The integrating factor of the differential eq | vedclass

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

Question

(B) Given differential equation is $(1+y+x^{2}y)dx + (x+x^{3})dy = 0$. Rearranging the terms,we have $(x+x^{3})dy = -(1+y(1+x^{2}))dx$. Dividing by $d

Vedclass · Accepted Answer

$x$

Vedclass · Answer

(B) Given differential equation is $(1+y+x^{2}y)dx + (x+x^{3})dy = 0$. Rearranging the terms,we have $(x+x^{3})dy = -(1+y(1+x^{2}))dx$. Dividing by $dx(x+x^{3})$,we get $\frac{dy}{dx} = -\frac{1+y(1+x^{2})}{x(1+x^{2})}$. This simplifies to $\frac{dy}{dx} = -\frac{1}{x(1+x^{2})} - \frac{y(1+x^{2})}{x(1+x^{2})}$. Thus,$\frac{dy}{dx} + \left(\frac{1}{x}\right)y = -\frac{1}{x(1+x^{2})}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = -\frac{1}{x(1+x^{2})}$. The integrating factor ($I$.$F$.) is given by $e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$.

The integrating factor of the differential equation $(1+y+x^{2}y)dx + (x+x^{3})dy = 0$ is

Similar Questions

If $y(t)$ is a solution of $(1 + t)\frac{dy}{dt} - ty = 1$ and $y(0) = -1$,then $y(1)$ is equal to

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 integrating factor of $\left(x+2 y^3\right) \frac{d y}{d x}=y^2$ is

The integrating factor of $y + \frac{d}{dx}(xy) = x(\sin x + \log x)$ is

Let $y = y(x)$ be the solution of the differential equation $\cos x \frac{dy}{dx} + 2y \sin x = \sin 2x$ for $x \in (0, \frac{\pi}{2})$. If $y(\frac{\pi}{3}) = 0$,then $y(\frac{\pi}{4})$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module