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

$A$ curve having the condition that the slope of the tangent at any point $(x, y)$ is two times the slope of the straight line joining the same point to the origin $(0, 0)$ is a/an:

Let $f: R \rightarrow R$ be a differentiable function with $f(0)=0$. If $y=f(x)$ satisfies the differential equation $\frac{dy}{dx}=(2+5y)(5y-2)$,then the value of $\lim_{x \rightarrow -\infty} f(x)$ is:

The solution of $(x - y^3)dx + 3xy^2dy = 0$ is

Find a particular solution of the differential equation $(x-y)(dx+dy)=dx-dy$ given that $y=-1$ when $x=0$. (Hint: put $x-y=t$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module