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 function $y=f(x)$ is the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}}$ in $(-1,1)$ satisfying $f(0)=0$. Then $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$ is

The general solution of the equation $\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{x}e^x$ is

If $\frac{dy}{dx} + y \tan x = \sin 2x$ and $y(0) = 1$,then $y(\pi)$ is equal to

The integrating factor of $x \frac{dy}{dx} - 2y = x^2 + \sin \left( \frac{1}{x^2} \right)$ is

Let $y=y(x), x>1$,be the solution of the differential equation $(x-1) \frac{d y}{d x}+2 x y=\frac{1}{x-1}$,with $y(2)=\frac{1+e^{4}}{2 e^{4}}$. If $y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}}$,then the value of $\alpha+\beta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module