Let y = y(x) be the solution of the different | vedclass

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

Question

(A) Given the differential equation: $\cos x \frac{dy}{dx} + 2y \sin x = \sin 2x$. Dividing by $\cos x$,we get: $\frac{dy}{dx} + 2y 	an x = 2 \sin x$

Vedclass · Accepted Answer

$\sqrt{2} - 2$

Vedclass · Answer

(A) Given the differential equation: $\cos x \frac{dy}{dx} + 2y \sin x = \sin 2x$. Dividing by $\cos x$,we get: $\frac{dy}{dx} + 2y 	an x = 2 \sin x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2 	an x$ and $Q = 2 \sin x$. The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int 2 	an x dx} = e^{2 \ln |\sec x|} = \sec^2 x$. The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$. $y \sec^2 x = \int 2 \sin x \cdot \sec^2 x dx + C$. $y \sec^2 x = 2 \int 	an x \sec x dx + C$. $y \sec^2 x = 2 \sec x + C$. Given $y(\frac{\pi}{3}) = 0$,we substitute $x = \frac{\pi}{3}$ and $y = 0$: $0 \cdot \sec^2(\frac{\pi}{3}) = 2 \sec(\frac{\pi}{3}) + C$. $0 = 2(2) + C \implies C = -4$. Thus,the solution is $y \sec^2 x = 2 \sec x - 4$. For $x = \frac{\pi}{4}$,$y \sec^2(\frac{\pi}{4}) = 2 \sec(\frac{\pi}{4}) - 4$. $y(2) = 2(\sqrt{2}) - 4$. $2y = 2\sqrt{2} - 4$. $y = \sqrt{2} - 2$.

Let $y = y(x)$ be the solution of the differential equation $\cos x \frac{dy}{dx} + 2y \sin x = \sin 2x$ for $x \in (0, \frac{\pi}{2})$. If $y(\frac{\pi}{3}) = 0$,then $y(\frac{\pi}{4})$ is equal to:

Similar Questions

The solution of $r dx + (x - r^2) dr = 0$ is

The solution of the differential equation $(x + 2y^3)\frac{dy}{dx} - y = 0$ is

Find the solution of the differential equation $(e^{y-x}) dy = (e^x - e^y) dx$.

If $y = \left(\frac{2}{\pi} x - 1\right) \operatorname{cosec} x$ is the solution of the differential equation $\frac{dy}{dx} + p(x) y = \frac{2}{\pi} \operatorname{cosec} x$ for $0 < x < \frac{\pi}{2}$,then the function $p(x)$ is equal to

Find the equation of a curve passing through the origin,given that the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the coordinates of the point.

Vedclass Test Series

Exam Paper Generator

Online Exam Module