The general solution of sin y cdot frac{dy}{d | vedclass

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

Question

(B) Given differential equation is $\sin y \frac{dy}{dx} = \cos y(1 - x \cos y)$.Dividing both sides by $\cos^2 y$,we get:$\frac{\sin y}{\cos^2 y} \fr

Vedclass · Accepted Answer

$\sec y = x + 1 + c e^x$

Vedclass · Answer

(B) Given differential equation is $\sin y \frac{dy}{dx} = \cos y(1 - x \cos y)$.Dividing both sides by $\cos^2 y$,we get:$\frac{\sin y}{\cos^2 y} \frac{dy}{dx} = \frac{1}{\cos y} - x$$\sec y 	an y \frac{dy}{dx} = \sec y - x$Let $\sec y = t$. Then $\sec y 	an y \frac{dy}{dx} = \frac{dt}{dx}$.Substituting this into the equation,we get the linear differential equation:$\frac{dt}{dx} = t - x \implies \frac{dt}{dx} - t = -x$.The integrating factor $(IF)$ is $e^{\int -1 dx} = e^{-x}$.The solution is $t(IF) = \int (-x)(IF) dx + c$.$t e^{-x} = \int -x e^{-x} dx + c$.Using integration by parts for $\int -x e^{-x} dx$:$\int -x e^{-x} dx = x e^{-x} - \int e^{-x} dx = x e^{-x} + e^{-x} + c$.Thus,$t e^{-x} = x e^{-x} + e^{-x} + c$.Multiplying by $e^x$,we get $t = x + 1 + c e^x$.Substituting back $t = \sec y$,we get $\sec y = x + 1 + c e^x$.

The general solution of $\sin y \cdot \frac{dy}{dx} = \cos y(1 - x \cos y)$ is

Similar Questions

If $y = A(x) e^{\int P dx}$ is a solution of $\frac{dy}{dx} + P(x) y = Q(x)$,then $A'(x) =$

The equation of one of the curves whose slope at any point is equal to $y+2x$ is

The integrating factor of the differential equation $\frac{dy}{dx} + (3x^2 \tan^{-1} y - x^3)(1 + y^2) = 0$ is

Let $f: R \rightarrow R$ be a continuous function satisfying $f(x) + \int_{0}^{x} t f(t) dt + x^2 = 0$ for all $x \in R$. Then:

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x$,$x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$,such that $y(0) = 1$. Then

Vedclass Test Series

Exam Paper Generator

Online Exam Module