Let the slope of the tangent to a curve y=f(x | vedclass

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

Question

(B) The slope of the tangent is given by $\frac{dy}{dx} = 2 	an x(\cos x - y)$.This can be rewritten as the linear differential equation: $\frac{dy}{

Vedclass · Accepted Answer

$2-\frac{\pi}{\sqrt{2}}$

Vedclass · Answer

(B) The slope of the tangent is given by $\frac{dy}{dx} = 2 	an x(\cos x - y)$.This can be rewritten as the linear differential equation: $\frac{dy}{dx} + (2 	an x)y = 2 \sin x$.The integrating factor is $IF = e^{\int 2 	an x \, dx} = e^{2 \ln |\sec x|} = \sec^2 x$.Multiplying both sides by $IF$,we get $\frac{d}{dx}(y \sec^2 x) = 2 \sin x \sec^2 x = 2 \sec x 	an x$.Integrating both sides,$y \sec^2 x = 2 \sec x + C$,which simplifies to $y = 2 \cos x + C \cos^2 x$.Since the curve passes through $(\frac{\pi}{4}, 0)$,we have $0 = 2 \cos(\frac{\pi}{4}) + C \cos^2(\frac{\pi}{4}) = 2(\frac{1}{\sqrt{2}}) + C(\frac{1}{2}) = \sqrt{2} + \frac{C}{2}$.Thus,$C = -2\sqrt{2}$,and the curve is $y = 2 \cos x - 2\sqrt{2} \cos^2 x$.Now,$\int_{0}^{\pi / 2} y \, dx = \int_{0}^{\pi / 2} (2 \cos x - 2\sqrt{2} \cos^2 x) \, dx$.$= [2 \sin x]_{0}^{\pi / 2} - 2\sqrt{2} \int_{0}^{\pi / 2} \frac{1 + \cos 2x}{2} \, dx$.$= 2(1 - 0) - \sqrt{2} [x + \frac{\sin 2x}{2}]_{0}^{\pi / 2} = 2 - \sqrt{2}(\frac{\pi}{2}) = 2 - \frac{\pi}{\sqrt{2}}$.

Let the slope of the tangent to a curve $y=f(x)$ at $(x, y)$ be given by $2 \tan x(\cos x-y)$. If the curve passes through the point $(\frac{\pi}{4}, 0)$,then the value of $\int_{0}^{\pi / 2} y \, dx$ is equal to

Similar Questions

The solution of the differential equation,$x^2 \frac{dy}{dx} \cos \frac{1}{x} - y \sin \frac{1}{x} = -1,$ where $y \rightarrow -1$ as $x \rightarrow \infty$ is

Let $y=y(x)$ be the solution curve of the differential equation $(1+x^{2})dy+(y-\tan^{-1}x)dx=0$,with $y(0)=1$. Then the value of $y(1)$ is:

Which of the following equations is a linear differential equation?

The integrating factor of the differential equation $\sin x \frac{dy}{dx} - y \cos x = 1$ is

The equation of the curve passing through the origin and satisfying the equation $(1 + {x^2})\frac{{dy}}{{dx}} + 2xy = 4{x^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module