By eliminating the arbitrary constants from y | vedclass

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

Question

(C) The given equation is $y=(a+b) \sin (x+c)-d e^{x+e+f}$.Let $A = (a+b)$ and $B = d e^{e+f}$. Then the equation becomes $y = A \sin(x+c) - B e^x$.Th

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The given equation is $y=(a+b) \sin (x+c)-d e^{x+e+f}$.Let $A = (a+b)$ and $B = d e^{e+f}$. Then the equation becomes $y = A \sin(x+c) - B e^x$.There are $3$ independent arbitrary constants: $A$,$c$,and $B$.The order of the differential equation obtained by eliminating $n$ independent arbitrary constants is $n$.Since there are $3$ independent arbitrary constants,the order of the differential equation is $3$.

By eliminating the arbitrary constants from $y=(a+b) \sin (x+c)-d e^{x+e+f}$,the differential equation obtained is of order

Similar Questions

The differential equation of the family of curves represented by the equation ${x^2}y = a$ is:

The differential equation corresponding to the family of parabolas whose axis is along $x=1$ is

The differential equation corresponding to the family of parabolas $y^2=4a(x+a)$,where $a$ is the parameter,is

The equation of any member of the family of all the ellipses whose axes are along the coordinate axes satisfies the differential equation

The differential equation of $3y = \sqrt[3]{x} + c$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module