Find the integrating factor of the differenti | vedclass

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

Question

(D) Given the differential equation: $(1+x^{2}) dt = (	an^{-1} x - t) dx$. Divide both sides by $(1+x^{2}) dx$: $\frac{dt}{dx} = \frac{	an^{-1} x -

Vedclass · Accepted Answer

$e^{	an^{-1} x}$

Vedclass · Answer

(D) Given the differential equation: $(1+x^{2}) dt = (	an^{-1} x - t) dx$. Divide both sides by $(1+x^{2}) dx$: $\frac{dt}{dx} = \frac{	an^{-1} x - t}{1+x^{2}}$ Rearrange the equation into the standard linear form $\frac{dt}{dx} + P(x)t = Q(x)$: $\frac{dt}{dx} + \frac{1}{1+x^{2}}t = \frac{	an^{-1} x}{1+x^{2}}$ Here,$P(x) = \frac{1}{1+x^{2}}$. The integrating factor ($I$.$F$.) is given by $e^{\int P(x) dx}$: $I.F. = e^{\int \frac{1}{1+x^{2}} dx} = e^{	an^{-1} x}$.

Find the integrating factor of the differential equation $(1+x^{2}) dt = (\tan^{-1} x - t) dx$.

Similar Questions

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=2(y+2 \sin x-5)x-2 \cos x$ such that $y(0)=7$. Then $y(\pi)$ is equal to :

The solution of the differential equation $\frac{dy}{dx} + \frac{y}{x} = x^2$ is

If for the solution curve $y=f(x)$ of the differential equation $\frac{dy}{dx}+(\tan x)y=\frac{2+\sec x}{(1+2\sec x)^2}$,$x \in \left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$,$f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}$,then $f\left(\frac{\pi}{4}\right)$ is equal to:

The general solution of a differential equation of the type $\frac{dx}{dy} + P_{1}x = Q_{1}$ is

The general solution of $x(x-1) \frac{dy}{dx} = x^3(2x-1) + (x-2)y$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module