Which of the following differential equations | vedclass

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

Question

(A) The given equation is:$y=c_{1} e^{x}+c_{2} e^{-x}$  $(1)$Differentiating with respect to $x$,we get:$\frac{d y}{d x}=c_{1} e^{x}-c_{2} e^{-x}$Agai

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given equation is:$y=c_{1} e^{x}+c_{2} e^{-x}$  $(1)$Differentiating with respect to $x$,we get:$\frac{d y}{d x}=c_{1} e^{x}-c_{2} e^{-x}$Again,differentiating with respect to $x$,we get:$\frac{d^{2} y}{d x^{2}}=c_{1} e^{x}+c_{2} e^{-x}$Since the right-hand side is equal to $y$ from equation $(1)$,we have:$\frac{d^{2} y}{d x^{2}}=y$Rearranging the terms,we get:$\frac{d^{2} y}{d x^{2}}-y=0$This is the required differential equation.Hence,the correct answer is $A$.

Which of the following differential equations has $y=c_{1} e^{x}+c_{2} e^{-x}$ as the general solution?

Similar Questions

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

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

Verify that the function $y=e^{-3x}$ is a solution of the differential equation $\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}-6y=0$.

Verify that the given function $y = \cos x + C$ is a solution of the differential equation $y^{\prime} + \sin x = 0$.

Find the differential equation for the family of curves given by eliminating arbitrary constants $a$ and $b$ from the equation: $\frac{x}{a} + \frac{y}{b} = 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module