The differential equation of y=e^x(a cos x+b | vedclass

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

Question

(D) Given equation: $y = e^x(a \cos x + b \sin x)$ Differentiating with respect to $x$ using the product rule: $\frac{dy}{dx} = e^x(a \cos x + b \sin

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given equation: $y = e^x(a \cos x + b \sin x)$ Differentiating with respect to $x$ using the product rule: $\frac{dy}{dx} = e^x(a \cos x + b \sin x) + e^x(-a \sin x + b \cos x)$ Since $y = e^x(a \cos x + b \sin x)$,we can write: $\frac{dy}{dx} = y + e^x(b \cos x - a \sin x)$ Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{dy}{dx} + [e^x(b \cos x - a \sin x) + e^x(-b \sin x - a \cos x)]$ Substitute $e^x(b \cos x - a \sin x) = \frac{dy}{dx} - y$: $\frac{d^2y}{dx^2} = \frac{dy}{dx} + (\frac{dy}{dx} - y) - e^x(a \cos x + b \sin x)$ Since $e^x(a \cos x + b \sin x) = y$,we have: $\frac{d^2y}{dx^2} = \frac{dy}{dx} + \frac{dy}{dx} - y - y$ $\frac{d^2y}{dx^2} = 2\frac{dy}{dx} - 2y$ Rearranging the terms: $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$

The differential equation of $y=e^x(a \cos x+b \sin x)$ is

Similar Questions

Form the differential equation representing the family of curves given by $(x-a)^{2}+2 y^{2}=a^{2},$ where $a$ is an arbitrary constant.

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

The differential equation of all the ellipses centered at the origin and having axes as the coordinate axes is

The differential equation formed by eliminating arbitrary constants $A$ and $B$ from the equation $y = A \cos 3x + B \sin 3x$ is

If $l$ and $m$ are the order and degree of the differential equation of all straight lines at a constant distance of $P$ units from the origin,then $l m^2+l^2 m=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module