If a, b, c, d are arbitrary constants,then th | vedclass

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

Question

(A) Given the equation: $y = a e^x + b e^{-x} + c \cos x + d \sin x$ Taking the first derivative: $y' = a e^x - b e^{-x} - c \sin x + d \cos x$ Taking

Vedclass · Accepted Answer

$y^{(4)} = y$

Vedclass · Answer

(A) Given the equation: $y = a e^x + b e^{-x} + c \cos x + d \sin x$ Taking the first derivative: $y' = a e^x - b e^{-x} - c \sin x + d \cos x$ Taking the second derivative: $y'' = a e^x + b e^{-x} - c \cos x - d \sin x$ Taking the third derivative: $y''' = a e^x - b e^{-x} + c \sin x - d \cos x$ Taking the fourth derivative: $y^{(4)} = a e^x + b e^{-x} + c \cos x + d \sin x$ Comparing the fourth derivative with the original equation,we get $y^{(4)} = y$,which can be written as $y^{(4)} - y = 0$.

If $a, b, c, d$ are arbitrary constants,then the corresponding differential equation for $y = a e^x + b e^{-x} + c \cos x + d \sin x$ is

Similar Questions

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

The family of curves $y = e^{a \sin x}$, where '$a$' is an arbitrary constant, is represented by the differential equation:

The order of the differential equation of the family of all concentric circles centered at $(h, k)$ is

Form the differential equation representing the family of curves $y = mx$,where $m$ is an arbitrary constant.

The differential equation for the line $y = mx + c$ is (where $m$ and $c$ are arbitrary constants).

Vedclass Test Series

Exam Paper Generator

Online Exam Module