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 for which $y = ax^2 + bx + c$ is the general solution is:

The differential equation of the family of circles passing through the origin and having their center on the line $y=x$ is:

The differential equation of the family of hyperbolas having their centres at the origin and their axes along the coordinate axes is

If the differential equation having $y=Ae^x+B \sin x$ as its general solution is $f(x) \frac{d^2 y}{d x^2}+g(x) \frac{d y}{d x}+h(x) y=0$,then $f(x)+g(x)+h(x)=$

Form the differential equation of the family of circles touching the $x$-axis at the origin.

Vedclass Test Series

Exam Paper Generator

Online Exam Module