Find a particular solution of the differentia | vedclass

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

Question

(A) The given differential equation is: $\frac{dy}{dx} + y \cot x = 4x \csc x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py

Vedclass · Accepted Answer

$y \sin x = 2x^2 - \frac{\pi^2}{2}$

Vedclass · Answer

(A) The given differential equation is: $\frac{dy}{dx} + y \cot x = 4x \csc x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cot x$ and $Q = 4x \csc x$. First,we find the integrating factor $(I.F.)$: $I.F. = e^{\int P dx} = e^{\int \cot x dx} = e^{\ln|\sin x|} = \sin x$. The general solution is given by $y(I.F.) = \int (Q \cdot I.F.) dx + C$. Substituting the values: $y \sin x = \int (4x \csc x \cdot \sin x) dx + C$. Since $\csc x \cdot \sin x = 1$,we have $y \sin x = \int 4x dx + C$. Integrating,we get $y \sin x = 2x^2 + C$. Given that $y=0$ when $x=\frac{\pi}{2}$,we substitute these values: $0 \cdot \sin(\frac{\pi}{2}) = 2(\frac{\pi}{2})^2 + C$. $0 = 2(\frac{\pi^2}{4}) + C \Rightarrow 0 = \frac{\pi^2}{2} + C \Rightarrow C = -\frac{\pi^2}{2}$. Wait,re-evaluating the constant: $0 = 2(\frac{\pi^2}{4}) + C \Rightarrow 0 = \frac{\pi^2}{2} + C \Rightarrow C = -\frac{\pi^2}{2}$. Thus,the particular solution is $y \sin x = 2x^2 - \frac{\pi^2}{2}$.

Find a particular solution of the differential equation $\frac{dy}{dx} + y \cot x = 4x \csc x$ $(x \neq 0),$ given that $y=0$ when $x=\frac{\pi}{2}$.

Similar Questions

Let $y=y(x)$ be the solution of the differential equation $x dy = (y + x^3 \cos x) dx$ with $y(\pi) = 0$. Then $y(\frac{\pi}{2})$ is equal to:

Find a particular solution satisfying the given condition: $\frac{dy}{dx} + 2y \tan x = \sin x$; $y = 0$ when $x = \frac{\pi}{3}$.

The general solution of the differential equation $\frac{d y}{d x}=\frac{2 y^2+1}{2 y^3-4 x y+y}$ is

The solution of the differential equation $\frac{dy}{dx} - 2y \tan 2x = e^x \sec 2x$ is

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 2y = f(x)$,where $f(x) = \begin{cases} 1, & x \in [0, 1] \\ 0, & \text{otherwise} \end{cases}$. If $y(0) = 0$,then $y\left(\frac{3}{2}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module