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

The solution of $dy = \cos x(2 - y \csc x)dx$ where $y = 2$ when $x = \frac{\pi}{2}$ is

For $x \in R$,let the function $y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 12y = \cos \left(\frac{\pi}{12} x\right)$ with $y(0) = 0$. Then,which of the following statements is/are $TRUE$?

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 3(\tan^2 x + 1)y = \sec^2 x$,with the initial condition $y(0) = \frac{1}{3} + e^3$. Then $y\left(\frac{\pi}{4}\right)$ is equal to:

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

Let $y=y(t)$ be a solution of the differential equation $\frac{dy}{dt}+\alpha y=\gamma e^{-\beta t}$,where $\alpha > 0, \beta > 0$ and $\gamma > 0$. Then $\lim_{t \rightarrow \infty} y(t)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module