The order of the differential equation obtain | vedclass

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

Question

(A) Given the equation: $c_{1} y = (c_{2} + c_{3}) e^{x + c_{4}}$We can simplify this by grouping the constants: $y = \left( \frac{c_{2} + c_{3}}{c_{1

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the equation: $c_{1} y = (c_{2} + c_{3}) e^{x + c_{4}}$We can simplify this by grouping the constants: $y = \left( \frac{c_{2} + c_{3}}{c_{1}} \right) e^{c_{4}} \cdot e^{x}$Let $C = \left( \frac{c_{2} + c_{3}}{c_{1}} \right) e^{c_{4}}$,where $C$ is a single arbitrary constant.Thus,the equation becomes $y = C e^{x}$.Since there is only one independent arbitrary constant $C$,the order of the differential equation obtained by eliminating it is $1$.Differentiating $y = C e^{x}$ with respect to $x$,we get $\frac{dy}{dx} = C e^{x}$.Substituting $y = C e^{x}$ into the derivative,we get $\frac{dy}{dx} = y$.This is a first-order differential equation,so the order is $1$.

The order of the differential equation obtained by eliminating arbitrary constants in the family of curves $c_{1} y = (c_{2} + c_{3}) e^{x + c_{4}}$ is

Similar Questions

The differential equation of the family of circles passing through the origin and having their centres on the $x$-axis is

The order of the differential equation of all parabolas whose axis of symmetry is along the $X$-axis is:

The order of the differential equation of all circles whose radius is $4$,is . . . . . .

Form the differential equation representing the family of curves given by $(x-a)^{2}+2 y^{2}=a^{2},$ where $a$ is an arbitrary constant.

If $m$ and $n$ are respectively the order and degree of the differential equation of the family of parabolas with focus at the origin and $X$-axis as its axis,then $m n-m+n=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module