If y=y(x) is the solution of frac{dy}{dx}=fra | vedclass

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

Question

(A) Given the differential equation $\frac{dy}{dx}=\frac{x-y \cos x}{1+\sin x}$.Rearranging the terms,we get $\frac{dy}{dx} + \left(\frac{\cos x}{1+\s

Vedclass · Accepted Answer

$\frac{5 \pi^2}{8}$

Vedclass · Answer

(A) Given the differential equation $\frac{dy}{dx}=\frac{x-y \cos x}{1+\sin x}$.Rearranging the terms,we get $\frac{dy}{dx} + \left(\frac{\cos x}{1+\sin x}\right)y = \frac{x}{1+\sin x}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{\cos x}{1+\sin x}$ and $Q(x) = \frac{x}{1+\sin x}$.The integrating factor $I.F. = e^{\int P(x) dx} = e^{\int \frac{\cos x}{1+\sin x} dx} = e^{\ln(1+\sin x)} = 1+\sin x$.The general solution is $y(I.F.) = \int Q(x)(I.F.) dx + C$.$y(1+\sin x) = \int \frac{x}{1+\sin x} (1+\sin x) dx + C = \int x dx + C = \frac{x^2}{2} + C$.Using the condition $y\left(\frac{\pi}{2}\right) = \frac{\pi^2}{8}$:$\frac{\pi^2}{8}(1+\sin(\frac{\pi}{2})) = \frac{(\pi/2)^2}{2} + C \Rightarrow \frac{\pi^2}{8}(2) = \frac{\pi^2}{8} + C \Rightarrow \frac{\pi^2}{4} - \frac{\pi^2}{8} = C \Rightarrow C = \frac{\pi^2}{8}$.Thus,$y(1+\sin x) = \frac{x^2}{2} + \frac{\pi^2}{8}$.For $x = \pi$,$y(1+\sin \pi) = \frac{\pi^2}{2} + \frac{\pi^2}{8}$.Since $\sin \pi = 0$,$y(1) = \frac{4\pi^2 + \pi^2}{8} = \frac{5\pi^2}{8}$.

If $y=y(x)$ is the solution of $\frac{dy}{dx}=\frac{x-y \cos x}{1+\sin x}$ with $y\left(\frac{\pi}{2}\right)=\frac{\pi^2}{8}$,then $y(\pi)=$

Similar Questions

An integrating factor for the differential equation $(1 + y^2)dx - (\tan^{-1} y - x)dy = 0$ is:

Find the general solution of the differential equation $y dx - (x + 2y^2) dy = 0$.

If $x = x(y)$ is the solution of the differential equation $y \frac{dx}{dy} = 2x + y^{3}(y+1)e^{y}$ with the initial condition $x(1) = 0$,then $x(e)$ is equal to:

Let $y=y(x)$ be the solution of the differential equation $(xy-5x^2\sqrt{1+x^2})dx+(1+x^2)dy=0$,with $y(0)=0$. Then $y(\sqrt{3})$ is equal to

The solution of $\frac{dy}{dx} + \frac{1}{3}y = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module