The differential equation of the family of cu | vedclass

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

Question

(A) Given $y=e^{x}(A \cos x+B \sin x)$ ... $(1)$Differentiating with respect to $x$:$\frac{dy}{dx} = e^{x}(A \cos x + B \sin x) + e^{x}(-A \sin x + B

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $y=e^{x}(A \cos x+B \sin x)$ ... $(1)$Differentiating with respect to $x$:$\frac{dy}{dx} = e^{x}(A \cos x + B \sin x) + e^{x}(-A \sin x + B \cos x)$$\frac{dy}{dx} = y + e^{x}(-A \sin x + B \cos x)$ ... $(2)$Differentiating again with respect to $x$:$\frac{d^{2}y}{dx^{2}} = \frac{dy}{dx} + e^{x}(-A \sin x + B \cos x) + e^{x}(-A \cos x - B \sin x)$Substitute $e^{x}(-A \sin x + B \cos x) = \frac{dy}{dx} - y$ from $(2)$:$\frac{d^{2}y}{dx^{2}} = \frac{dy}{dx} + (\frac{dy}{dx} - y) - e^{x}(A \cos x + B \sin x)$Since $e^{x}(A \cos x + B \sin x) = y$:$\frac{d^{2}y}{dx^{2}} = 2\frac{dy}{dx} - y - y$$\frac{d^{2}y}{dx^{2}} - 2\frac{dy}{dx} + 2y = 0$

The differential equation of the family of curves $y=e^{x}(A \cos x+B \sin x)$,where $A$ and $B$ are arbitrary constants,is

Similar Questions

If $m$ and $n$ are respectively the order and degree of the differential equation of the family of parabolas with focus at the origin and $X$-axis as its axis,then $m n-m+n=$

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

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

$y=A e^x+B e^{2 x}+C e^{3 x}$ satisfies the differential equation

The differential equation formed by eliminating $a$ and $b$ from the equation $y=a e^{2 x}+b x e^{2 x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module