General solution of the differential equation | vedclass

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

Question

(B) Given the differential equation $\sin^3 x \frac{dx}{dy} = \sin y$. Separating the variables,we get $\int \sin^3 x \, dx = \int \sin y \, dy$. Usin

Vedclass · Accepted Answer

$\cos y - \frac{3}{4} \cos x + \frac{1}{12} \cos 3x = C$

Vedclass · Answer

(B) Given the differential equation $\sin^3 x \frac{dx}{dy} = \sin y$. Separating the variables,we get $\int \sin^3 x \, dx = \int \sin y \, dy$. Using the trigonometric identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we have $\sin^3 x = \frac{3 \sin x - \sin 3x}{4}$. Substituting this into the integral,we get $\int \frac{3 \sin x - \sin 3x}{4} \, dx = \int \sin y \, dy$. Integrating both sides,we get $\frac{3}{4} (-\cos x) - \frac{1}{4} (-\frac{\cos 3x}{3}) = -\cos y + C$. This simplifies to $-\frac{3}{4} \cos x + \frac{1}{12} \cos 3x = -\cos y + C$. Rearranging the terms,we get $\cos y - \frac{3}{4} \cos x + \frac{1}{12} \cos 3x = C$.

General solution of the differential equation $\sin^3 x \frac{dx}{dy} = \sin y$ is given by

Similar Questions

The general solution of the differential equation $\tan(y) dx + \sec^2(y) \tan(x) dy = 0$ is

Let $y=y(x)$ be the solution of the differential equation $e^{x} \sqrt{1-y^{2}} dx + \frac{y}{x} dy = 0$,with $y(1) = -1$. Then the value of $(y(3))^{2}$ is equal to:

If $dx + dy = (x + y)(dx - dy)$, then $\log(x + y)$ is equal to

The particular solution of the differential equation $\frac{dy}{dx} - e^x = y e^x$,when $x = 0$ and $y = 1$ is

The solution of $\frac{dy}{dx} = \left( \frac{y}{x} \right)^{1/3}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module