The differential equation x frac{dy}{dx} = 2y | vedclass

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

Question

(B) Given the differential equation: $x \frac{dy}{dx} = 2y$. Separating the variables,we get: $\frac{dy}{y} = 2 \frac{dx}{x}$. Integrating both sides:

Vedclass · Accepted Answer

a family of parabolas with vertex at the origin and axis along the positive $Y$-axis.

Vedclass · Answer

(B) Given the differential equation: $x \frac{dy}{dx} = 2y$. Separating the variables,we get: $\frac{dy}{y} = 2 \frac{dx}{x}$. Integrating both sides: $\int \frac{dy}{y} = 2 \int \frac{dx}{x}$. This gives: $\ln|y| = 2 \ln|x| + C$. Using properties of logarithms: $\ln|y| = \ln|x^2| + C$. Taking the exponential of both sides: $y = e^{\ln|x^2| + C} = e^C \cdot x^2$. Let $e^C = k$,where $k$ is an arbitrary constant. Thus,$y = kx^2$. This equation represents a family of parabolas with the vertex at the origin $(0,0)$ and the axis along the $Y$-axis.

The differential equation $x \frac{dy}{dx} = 2y$ represents ....

Similar Questions

The slope of the tangent to a curve $C : y = y(x)$ at any point $(x, y)$ on it is $\frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}$. If $C$ passes through the points $(0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}})$ and $(\alpha, \frac{1}{2}e^{2\alpha})$,then $e^{\alpha}$ is equal to:

If the tangent at a point $P(x, y)$ of a curve is perpendicular to the line that joins the origin with the point $P$,then the curve is

$A$ radioactive substance,with initial mass $m_0$,has a half-life of $h$ days. Then its initial decay rate is given by

The family of curves whose $x$ and $y$ intercepts of a tangent at any point are respectively double the $x$ and $y$ coordinates of that point is

The curve passing through $(3, 0)$ and satisfying the differential equation $(9 - x^2)(\frac{dy}{dx})^2 = 9 - y^2$ represents:

Vedclass Test Series

Exam Paper Generator

Online Exam Module