The general solution of the differential equa | vedclass

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

Question

(D) Given,$\sin 2x \left( \frac{dy}{dx} - \sqrt{	an x} ight) - y = 0$ Rearranging the terms: $\frac{dy}{dx} - \frac{y}{\sin 2x} = \sqrt{	an x}$ Si

Vedclass · Accepted Answer

$y\sqrt{\cot x} = x + c$

Vedclass · Answer

(D) Given,$\sin 2x \left( \frac{dy}{dx} - \sqrt{	an x} ight) - y = 0$ Rearranging the terms: $\frac{dy}{dx} - \frac{y}{\sin 2x} = \sqrt{	an x}$ Since $\frac{1}{\sin 2x} = \csc 2x$,we have: $\frac{dy}{dx} - y \csc 2x = \sqrt{	an x}$ ....$(1)$ This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\csc 2x$ and $Q = \sqrt{	an x}$. Integrating Factor ($I$.$F$.) $= e^{\int P dx} = e^{\int -\csc 2x dx} = e^{-\frac{1}{2} \ln|	an x|} = e^{\ln(	an x)^{-1/2}} = \frac{1}{\sqrt{	an x}} = \sqrt{\cot x}$. The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$. Substituting the values: $y \sqrt{\cot x} = \int \sqrt{	an x} \cdot \sqrt{\cot x} dx + c$. Since $\sqrt{	an x} \cdot \sqrt{\cot x} = 1$,we get: $y \sqrt{\cot x} = \int 1 dx + c$. Therefore,$y \sqrt{\cot x} = x + c$.

The general solution of the differential equation,$\sin 2x \left( \frac{dy}{dx} - \sqrt{\tan x} \right) - y = 0$ is

Similar Questions

Let the solution $y=y(x)$ of the differential equation $\frac{dy}{dx}-y=1+4 \sin x$ satisfy $y(\pi)=1$. Then $y\left(\frac{\pi}{2}\right)+10$ is equal to:

The solution of the differential equation $(1+x) \frac{dy}{dx} - xy = 1-x$ is

The solution of the differential equation $x\frac{dy}{dx} + y = x^2 + 3x + 2$ is

If $x=f(y)$ is the solution of the differential equation $(1+y^2)+(x-2 e^{\tan ^{-1} y}) \frac{d y}{d x}=0$,$y \in(-\frac{\pi}{2}, \frac{\pi}{2})$ with $f(0)=1$,then $f(\frac{1}{\sqrt{3}})$ is equal to :

The general solution of the differential equation $x \sin x \frac{dy}{dx} + (x \cos x + \sin x)y = \sin x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module