The curve y=(cos x+y)^{1 / 2} satisfies the d | vedclass

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

Question

(A) Given curve is $y=(\cos x+y)^{1/2}$.Squaring both sides,we get $y^2 = \cos x + y$.Differentiating both sides with respect to $x$:$2y \frac{dy}{dx}

Vedclass · Accepted Answer

$(2 y-1) \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}+\cos x=0$

Vedclass · Answer

(A) Given curve is $y=(\cos x+y)^{1/2}$.Squaring both sides,we get $y^2 = \cos x + y$.Differentiating both sides with respect to $x$:$2y \frac{dy}{dx} = -\sin x + \frac{dy}{dx}$.Rearranging the terms:$(2y - 1) \frac{dy}{dx} = -\sin x$.Differentiating again with respect to $x$ using the product rule:$(2y - 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{d}{dx}(2y - 1) = -\cos x$.$(2y - 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot (2 \frac{dy}{dx}) = -\cos x$.$(2y - 1) \frac{d^2y}{dx^2} + 2 \left(\frac{dy}{dx}\right)^2 + \cos x = 0$.

The curve $y=(\cos x+y)^{1 / 2}$ satisfies the differential equation

Similar Questions

The differential equation of the family of circles passing through $(0,0)$ and having centre on the $X$-axis is

If $a$ and $b$ are arbitrary constants,then the differential equation having $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as its general solution is

The differential equation formed by eliminating $A$ and $B$ from $A x^2 + B y^2 = 1$ is

The differential equation corresponding to the primitive $y = e^{mx}$ is obtained by eliminating the arbitrary constant $m$. What is the resulting differential equation?

The elimination of the arbitrary constants $A, B$ and $C$ from $y = A + Bx + C{e^{ - x}}$ leads to the differential equation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module