The differential equation whose solution represents the family $x^2 y = 4e^x + c$,where $c$ is an arbitrary constant,is
- A$x \frac{dy}{dx} + xy = 0$
- B$x^2 \frac{dy}{dx} + (2xy - 4e^x) = 0$
- C$x \frac{dy}{dx} + (x - 2)y = 0$
- D$x \frac{dy}{dx} + (2 - x)y = 0$
Explore More
Similar Questions
The differential equation of the family of ellipses $\frac{x^2}{a^2} + \frac{y^2}{b^2} = c$ is given by $\left( y' = \frac{dy}{dx}, y'' = \frac{d^2y}{dx^2} \right)$.
Form the differential equation of the family of circles having centre on the $y$-axis and radius $3$ units.
Medium
View SolutionThe differential equation of the family of curves $y=e^{x}(A \cos x+B \sin x)$ where $A, B$ are arbitrary constants is
The degree of the differential equation obtained from the equation $(y - a)^2 = 4(x - b)$ [where $a$ and $b$ are arbitrary constants] is
DifficultMHT CET 2026
View SolutionThe differential equation of the family of curves $y = a e^x + b x e^x + c x^2 e^x$,where $a, b, c$ are arbitrary constants,is
DifficultTS EAMCET 2009
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo