y = a{e^{mx}} + b{e^{ - mx}} satisfies which | vedclass

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

Question

(D) Given the equation: $y = a{e^{mx}} + b{e^{-mx}}$Differentiating with respect to $x$,we get:$\frac{dy}{dx} = ma{e^{mx}} - mb{e^{-mx}}$Differentiati

Vedclass · Accepted Answer

$\frac{d^2y}{dx^2} - m^2y = 0$

Vedclass · Answer

(D) Given the equation: $y = a{e^{mx}} + b{e^{-mx}}$Differentiating with respect to $x$,we get:$\frac{dy}{dx} = ma{e^{mx}} - mb{e^{-mx}}$Differentiating again with respect to $x$,we get:$\frac{d^2y}{dx^2} = m^2a{e^{mx}} + m^2b{e^{-mx}}$Factoring out $m^2$:$\frac{d^2y}{dx^2} = m^2(a{e^{mx}} + b{e^{-mx}})$Since $y = a{e^{mx}} + b{e^{-mx}}$,we can substitute $y$ into the equation:$\frac{d^2y}{dx^2} = m^2y$Rearranging the terms gives:$\frac{d^2y}{dx^2} - m^2y = 0$Thus,the correct option is $D$.

$y = a{e^{mx}} + b{e^{ - mx}}$ satisfies which of the following differential equations?

Similar Questions

The differential equation of all circles having their centres on the line $y=5$ and touching the $X$-axis is ......

Form the differential equation representing the family of curves $y=a \sin (x+b),$ where $a$ and $b$ are arbitrary constants.

The differential equation of all circles having center on the $x$-axis and touching the $y$-axis at the origin is:

$y = e^x (A \cos x + B \sin x)$ is the solution of the differential equation

Find the differential equation for the family of curves $y^{2}=a(b^{2}-x^{2})$ by eliminating the arbitrary constants $a$ and $b$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module