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 having the general solution $y=c(x-c)^2$ ($c$ is an arbitrary constant) is

The order of the differential equation of all parabolas,whose latus rectum is $4a$ and axis is parallel to the $x$-axis,is

If the degree of the differential equation corresponding to the family of curves $y=ax+\frac{1}{a}$ (where $a \neq 0$ is an arbitrary constant) is $r$ and its order is $m$,then the solution of $\frac{dy}{dx}=\frac{y}{2x}, y(1)=\sqrt{r+m}$ is

The differential equation corresponding to the family of curves $y=e^x(A \cos x+B \sin x)$ is

The differential equation of all parabolas,whose axes are parallel to the $Y$-axis,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module