The differential equation satisfied by the fu | vedclass

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

Question

(D) Given the function $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}}$.Since the series is infinite,we can write it as $y = \sqrt{\

Vedclass · Accepted Answer

$(2y - 1)\frac{dy}{dx} - \cos x = 0$

Vedclass · Answer

(D) Given the function $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}}$.Since the series is infinite,we can write it as $y = \sqrt{\sin x + y}$.Squaring both sides,we get $y^2 = \sin x + y$.Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(y^2) = \frac{d}{dx}(\sin x + y)$.Using the chain rule,$2y \frac{dy}{dx} = \cos x + \frac{dy}{dx}$.Rearranging the terms,we get $2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x$.Factoring out $\frac{dy}{dx}$,we get $(2y - 1)\frac{dy}{dx} = \cos x$,which can be written as $(2y - 1)\frac{dy}{dx} - \cos x = 0$.

The differential equation satisfied by the function $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}}$ is

Similar Questions

The differential equation of all parabolas whose axis is the $y$-axis is:

The differential equation representing the family of parabolas having vertex at the origin and axis along the positive $Y$-axis 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 $f(x)+g(x)+h(x)=$

The differential equation representing the family of circles having their centres on the $Y$-axis is (where $y_1 = \frac{dy}{dx}$ and $y_2 = \frac{d^2y}{dx^2}$):

The general solution of the differential equation of all circles having center at $A(-1, 2)$ is $ . . . . . . $.

Vedclass Test Series

Exam Paper Generator

Online Exam Module