The solution of the differential equation 2x | vedclass

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

Question

(C) Given the differential equation: $2x \frac{dy}{dx} - y = 0$. Rearranging the terms,we get: $2x \frac{dy}{dx} = y$. Separating the variables: $\fra

Vedclass · Accepted Answer

$A$ parabola

Vedclass · Answer

(C) Given the differential equation: $2x \frac{dy}{dx} - y = 0$. Rearranging the terms,we get: $2x \frac{dy}{dx} = y$. Separating the variables: $\frac{dy}{y} = \frac{dx}{2x}$. Integrating both sides: $\int \frac{dy}{y} = \frac{1}{2} \int \frac{dx}{x}$. This gives: $\ln|y| = \frac{1}{2} \ln|x| + C$. Using the condition $y(1) = 2$: $\ln(2) = \frac{1}{2} \ln(1) + C \implies C = \ln(2)$. Substituting $C$ back: $\ln(y) = \ln(\sqrt{x}) + \ln(2) = \ln(2\sqrt{x})$. Thus,$y = 2\sqrt{x}$,which implies $y^2 = 4x$. The equation $y^2 = 4ax$ represents a parabola. Therefore,the correct option is $C$.

The solution of the differential equation $2x \frac{dy}{dx} - y = 0$ with the condition $y(1) = 2$ represents . . . . . . .

Similar Questions

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

Solve the differential equation: $\frac{dy}{dx} = e^{x+y}$

Find a particular solution satisfying the given condition: $\frac{dy}{dx} = y \tan x$; $y = 1$ when $x = 0$.

The general solution of the differential equation $\tan x \tan y \, dx + \cos^2 x \operatorname{cosec}^2 y \, dy = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module