The general solution of left(x frac{dy}{dx} - | vedclass

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

Question

(A) Given equation: $\left(x \frac{dy}{dx} - y\right) \sin \frac{y}{x} = x^3 e^x$ Divide both sides by $x^2$: $\left(\frac{x \frac{dy}{dx} - y}{x^2}\r

Vedclass · Accepted Answer

$e^x(x - 1) + \cos \frac{y}{x} + c = 0$

Vedclass · Answer

(A) Given equation: $\left(x \frac{dy}{dx} - y\right) \sin \frac{y}{x} = x^3 e^x$ Divide both sides by $x^2$: $\left(\frac{x \frac{dy}{dx} - y}{x^2}\right) \sin \frac{y}{x} = x e^x$ Let $t = \frac{y}{x}$. Then $\frac{dt}{dx} = \frac{x \frac{dy}{dx} - y}{x^2}$. Substituting this into the equation: $\frac{dt}{dx} \sin t = x e^x$ Integrating both sides with respect to $x$: $\int \sin t \, dt = \int x e^x \, dx$ Using integration by parts for the right side: $-\cos t = x e^x - \int e^x \, dx$ $-\cos t = x e^x - e^x + c$ $-\cos t = e^x(x - 1) + c$ Substituting $t = \frac{y}{x}$ back: $-\cos \frac{y}{x} = e^x(x - 1) + c$ Rearranging gives: $e^x(x - 1) + \cos \frac{y}{x} + c = 0$

The general solution of $\left(x \frac{dy}{dx} - y\right) \sin \frac{y}{x} = x^3 e^x$ is

Similar Questions

The general solution of the differential equation $\cos x \sin y \, dx + \sin x \cos y \, dy = 0$ is

The equation of the curve passing through $\left(\frac{\pi}{6}, 0\right)$ and satisfying the differential equation $(e^y+1) \cos x \, dx + e^y \sin x \, dy = 0$ is:

The solution of the differential equation $\frac{dy}{dx} = x^2 + \sin 3x$ is

$A$ particular solution of $\frac{dy}{dx} = (x + 9y)^2$,when $x = 0, y = \frac{1}{27}$ is

The particular solution of the differential equation $\cos \left(\frac{dy}{dx}\right) = a$,under the conditions $a \in \mathbb{R}$ and $y(0) = 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module