Find the particular solution of the different | vedclass

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

Question

(A) The given equation is a linear differential equation of the type $\frac{dy}{dx} + Py = Q$,where $P = \cot x$ and $Q = 2x + x^2 \cot x$.First,calcu

Vedclass · Accepted Answer

$y = x^2 - \frac{\pi^2}{4 \sin x}$

Vedclass · Answer

(A) The given equation is a linear differential equation of the type $\frac{dy}{dx} + Py = Q$,where $P = \cot x$ and $Q = 2x + x^2 \cot x$.First,calculate the Integrating Factor ($I$.$F$.):$I.F. = e^{\int \cot x dx} = e^{\log |\sin x|} = \sin x$.The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.$y \sin x = \int (2x + x^2 \cot x) \sin x dx + C$$y \sin x = \int 2x \sin x dx + \int x^2 \cos x dx + C$.Using integration by parts for $\int 2x \sin x dx$:Let $u = 2x, dv = \sin x dx \implies du = 2 dx, v = -\cos x$.$\int 2x \sin x dx = -2x \cos x - \int (-2 \cos x) dx = -2x \cos x + 2 \sin x$.Substituting this back:$y \sin x = (-2x \cos x + 2 \sin x) + \int x^2 \cos x dx + C$. Wait,let's use the simpler method:$y \sin x = \int (2x \sin x + x^2 \cos x) dx + C$.Note that $\frac{d}{dx}(x^2 \sin x) = 2x \sin x + x^2 \cos x$.Therefore,$y \sin x = x^2 \sin x + C$ $(1)$.Given $y = 0$ when $x = \frac{\pi}{2}$:$0 = (\frac{\pi}{2})^2 \sin(\frac{\pi}{2}) + C \implies 0 = \frac{\pi^2}{4} + C \implies C = -\frac{\pi^2}{4}$.Substituting $C$ into $(1)$:$y \sin x = x^2 \sin x - \frac{\pi^2}{4}$.Dividing by $\sin x$:$y = x^2 - \frac{\pi^2}{4 \sin x}$.

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

Similar Questions

Let $y=y(x)$ be the solution of the differential equation $\sec x \, dy + \{2(1-x) \tan x + x(2-x)\} \, dx = 0$ such that $y(0)=2$. Then $y(2)$ is equal to :

If $y=y(x)$ is the solution of $\frac{dy}{dx}=\frac{x-y \cos x}{1+\sin x}$ with $y\left(\frac{\pi}{2}\right)=\frac{\pi^2}{8}$,then $y(\pi)=$

Find the general solution of the differential equation: $(1+x^{2}) dy + 2xy dx = \cot x dx$ where $x \neq 0$.

The solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ $(x \neq 0)$ with $y(1) = 1$ is

The solution of the differential equation $y' = y \tan x - 2 \sin x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module