The particular solution of the differential e | vedclass

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

Question

(A) Given the differential equation: $\sin^{2} y \frac{dx}{dy} + x = \cot y$. Dividing by $\sin^{2} y$,we get: $\frac{dx}{dy} + (\operatorname{cosec}^

Vedclass · Accepted Answer

$x = 1 + \cot y$

Vedclass · Answer

(A) Given the differential equation: $\sin^{2} y \frac{dx}{dy} + x = \cot y$. Dividing by $\sin^{2} y$,we get: $\frac{dx}{dy} + (\operatorname{cosec}^{2} y)x = \cot y \operatorname{cosec}^{2} y$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \operatorname{cosec}^{2} y$ and $Q(y) = \cot y \operatorname{cosec}^{2} y$. The Integrating Factor ($I$.$F$.) is $e^{\int P(y) dy} = e^{\int \operatorname{cosec}^{2} y dy} = e^{-\cot y}$. The solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C$. $x e^{-\cot y} = \int \cot y \operatorname{cosec}^{2} y e^{-\cot y} dy + C$. Let $t = -\cot y$,then $dt = \operatorname{cosec}^{2} y dy$. $x e^{-\cot y} = \int (-t) e^{t} dt + C = -(t e^{t} - e^{t}) + C = e^{t}(1 - t) + C$. Substituting back $t = -\cot y$: $x e^{-\cot y} = e^{-\cot y}(1 + \cot y) + C$. Given $x = 0$ at $y = \frac{3\pi}{4}$,we have $0 = e^{-\cot(3\pi/4)}(1 + \cot(3\pi/4)) + C$. Since $\cot(3\pi/4) = -1$,$0 = e^{1}(1 - 1) + C \implies C = 0$. Thus,$x e^{-\cot y} = e^{-\cot y}(1 + \cot y)$,which simplifies to $x = 1 + \cot y$.

The particular solution of the differential equation $\sin^{2} y \frac{dx}{dy} + x = \cot y$ when $x = 0$ and $y = \frac{3\pi}{4}$ is

Similar Questions

Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function. If $10 \int_1^{x} f(t) dt = 5x f(x) - x^5 - 9$ for all $x \geq 1$,then the value of $f(3)$ is:

If the solution of the differential equation $(1 + x^3) \frac{dy}{dx} + 6x^2y = 1 + x^2$ is $y = \frac{1}{(1 + x^3)^s} [x + \frac{x^p}{p} + \frac{x^q}{q} + \frac{x^r}{r} + c]$, then the $LCM$ of $p, q, r$ and $s$ is...

The solution of the differential equation $x^2 \frac{dy}{dx} - xy = 1$ is...

Let $y=y(x), y>0$,be a solution curve of the differential equation $(1+x^2) dy = y(x-y) dx$. If $y(0)=1$ and $y(2\sqrt{2})=\beta$,then

General solution of the differential equation $\frac{dy}{dx} + y \tan x = \sec x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module