The differential equation frac{dy}{dx} + frac | vedclass

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

Question

(B) Given the differential equation: $\frac{dy}{dx} + \frac{1}{x} \sin 2y = x^3 \cos^2 y$. Using $\sin 2y = 2 \sin y \cos y$,we have $\frac{dy}{dx} +

Vedclass · Accepted Answer

$6x^2 	an y = x^6 + C$

Vedclass · Answer

(B) Given the differential equation: $\frac{dy}{dx} + \frac{1}{x} \sin 2y = x^3 \cos^2 y$. Using $\sin 2y = 2 \sin y \cos y$,we have $\frac{dy}{dx} + \frac{2}{x} \sin y \cos y = x^3 \cos^2 y$. Dividing both sides by $\cos^2 y$,we get: $\sec^2 y \frac{dy}{dx} + \frac{2}{x} 	an y = x^3$. Let $z = 	an y$,then $\frac{dz}{dx} = \sec^2 y \frac{dy}{dx}$. The equation becomes: $\frac{dz}{dx} + \frac{2}{x} z = x^3$. This is a linear differential equation of the form $\frac{dz}{dx} + P(x)z = Q(x)$,where $P(x) = \frac{2}{x}$ and $Q(x) = x^3$. The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln x} = x^2$. The solution is $z \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$. $z \cdot x^2 = \int x^3 \cdot x^2 dx + C = \int x^5 dx + C$. $z x^2 = \frac{x^6}{6} + C$. Substituting $z = 	an y$: $(	an y) x^2 = \frac{x^6}{6} + C$. Multiplying by $6$: $6x^2 	an y = x^6 + 6C$. Since $6C$ is an arbitrary constant,we can write it as $C$. Thus,$6x^2 	an y = x^6 + C$.

The differential equation $\frac{dy}{dx} + \frac{1}{x} \sin 2y = x^3 \cos^2 y$ represents a family of curves given by:

Similar Questions

The integrating factor of the differential equation $x \frac{dy}{dx} + 2y = x^2 \log x$ is

If $x \log x \frac{dy}{dx} + y = \log x^2$ and $y(e) = 0$,then $y(e^2) = $

If $\cos x$ and $\sin x$ are solutions of the differential equation $a_{0} \frac{d^{2} y}{d x^{2}}+a_{1} \frac{d y}{d x}+a_{2} y=0$ where $a_{0}, a_{1}$ and $a_{2}$ are real constants, then which of the following is/are always true?

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx} + \frac{\sqrt{2}y}{2\cos^4 x - \cos 2x} = x e^{\tan^{-1}(\sqrt{2} \cot 2x)}$,$0 < x < \pi/2$ with $y(\pi/4) = \pi^2/32$. If $y(\pi/3) = \frac{\pi^2}{18} e^{-\tan^{-1}(\alpha)}$,then the value of $3\alpha^2$ is equal to

The integrating factor of $x \frac{dy}{dx} + 3y = x^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module