Find the general solution of the differential | vedclass

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

Question

(A) Given equation: $x \frac{dy}{dx} + y - x + xy \cot x = 0$Divide by $x$: $\frac{dy}{dx} + \frac{y}{x} + y \cot x = 1$$\frac{dy}{dx} + y(\frac{1}{x}

Vedclass · Accepted Answer

$y = -\cot x + \frac{1}{x} + \frac{C}{x \sin x}$

Vedclass · Answer

(A) Given equation: $x \frac{dy}{dx} + y - x + xy \cot x = 0$Divide by $x$: $\frac{dy}{dx} + \frac{y}{x} + y \cot x = 1$$\frac{dy}{dx} + y(\frac{1}{x} + \cot x) = 1$This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x} + \cot x$ and $Q = 1$.Integrating Factor $(I.F.)$ $= e^{\int P dx} = e^{\int (\frac{1}{x} + \cot x) dx} = e^{\ln|x| + \ln|\sin x|} = e^{\ln|x \sin x|} = x \sin x$.The general solution is $y(I.F.) = \int (Q \cdot I.F.) dx + C$.$y(x \sin x) = \int (1 \cdot x \sin x) dx + C$.Using integration by parts: $\int x \sin x dx = x(-\cos x) - \int (1)(-\cos x) dx = -x \cos x + \sin x$.So,$y(x \sin x) = -x \cos x + \sin x + C$.Dividing by $x \sin x$: $y = \frac{-x \cos x}{x \sin x} + \frac{\sin x}{x \sin x} + \frac{C}{x \sin x}$.$y = -\cot x + \frac{1}{x} + \frac{C}{x \sin x}$.

Find the general solution of the differential equation: $x \frac{dy}{dx} + y - x + xy \cot x = 0$ $(x \neq 0)$.

Similar Questions

Integrating factor of the differential equation $\frac{dy}{dx} + y = \frac{1+y}{x}$ is

The general solution of the differential equation $(y^2 - x^3) dx - xy dy = 0, (x \neq 0)$ is (where $c$ is a constant of integration)

The general solution of the differential equation $\frac{dy}{dx} + \frac{y \ln y}{x} = \frac{y(\ln y)^2}{x^2}$ is (where $C$ is an arbitrary constant):

Let $f(x) = \int_{0}^{x} e^{t} f(t) dt + e^{x}$ be a differentiable function for all $x \in R$. Then $f(x)$ equals ..... .

Let the solution curve $y = y(x)$ of the differential equation $\frac{dy}{dx} - \frac{3x^5 \tan^{-1}(x^3)}{(1+x^6)^{3/2}} y = 2x \exp \left( \frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}} \right)$ pass through the origin. Then $y(1)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module