The differential equation whose solution is y | vedclass

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

Question

(B) Given the equation $y = cx + c - c^3$. To find the differential equation,we differentiate the given equation with respect to $x$: $\frac{dy}{dx} =

Vedclass · Accepted Answer

$y = x\frac{dy}{dx} + \frac{dy}{dx} - \left(\frac{dy}{dx}\right)^3$

Vedclass · Answer

(B) Given the equation $y = cx + c - c^3$. To find the differential equation,we differentiate the given equation with respect to $x$: $\frac{dy}{dx} = c$. Now,we substitute $c = \frac{dy}{dx}$ back into the original equation $y = cx + c - c^3$: $y = x\left(\frac{dy}{dx}\right) + \frac{dy}{dx} - \left(\frac{dy}{dx}\right)^3$. Thus,the required differential equation is $y = x\frac{dy}{dx} + \frac{dy}{dx} - \left(\frac{dy}{dx}\right)^3$.

The differential equation whose solution is $y = cx + c - c^3$ is:

Similar Questions

The differential equation corresponding to the family of curves given by $y=a+b e^{2 x}+c e^{-3 x}$ is

The equation which represents the system of parabolas whose axis is parallel to the $y$-axis satisfies the differential equation:

The differential equation of all straight lines passing through the origin is

The differential equation whose solution is $y = c_1 \cos ax + c_2 \sin ax$ is (where $c_1, c_2$ are arbitrary constants):

The differential equation whose solution is $Ax^2 + By^2 = 1$,where $A$ and $B$ are arbitrary constants,is of

Vedclass Test Series

Exam Paper Generator

Online Exam Module