The equation of the curve whose slope at any | vedclass

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

Question

(A) Given the slope of the curve is $\frac{dy}{dx} = 2xy$.Separating the variables,we get $\frac{dy}{y} = 2x \, dx$.Integrating both sides,we have $\i

Vedclass · Accepted Answer

$\log y = x^{2}$

Vedclass · Answer

(A) Given the slope of the curve is $\frac{dy}{dx} = 2xy$.Separating the variables,we get $\frac{dy}{y} = 2x \, dx$.Integrating both sides,we have $\int \frac{dy}{y} = \int 2x \, dx$.This gives $\log y = x^{2} + C$.Since the curve passes through the point $(0,1)$,we substitute $x = 0$ and $y = 1$ into the equation:$\log(1) = (0)^{2} + C \implies 0 = 0 + C \implies C = 0$.Therefore,the equation of the curve is $\log y = x^{2}$.

The equation of the curve whose slope at any point is equal to $2xy$ and which passes through the point $(0,1)$ is

Similar Questions

The general solution of the differential equation $(2x-y)^2 dy - 2(2x-y)^2 dx - 2 dx = 0$ is

The general solution of the differential equation $\log \left( \frac{dy}{dx} \right) = x + y$ is

The general solution of the differential equation $\frac{dy}{dx} + \sin \left( \frac{x + y}{2} \right) = \sin \left( \frac{x - y}{2} \right)$ is

If $c$ is any arbitrary constant,then the general solution of the differential equation $ydx - xdy = xy\,dx$ is given by

The general solution of the equation $({e^y} + 1)\cos x \, dx + {e^y}\sin x \, dy = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module