If y=f(x) is the solution of the differential | vedclass

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

Question

(B) The given differential equation is $(1+\cos^2 x) f'(x) - f(x) \sin 2x = 4 \sin 2x$. Divide by $(1+\cos^2 x)$ to get the linear form $f'(x) - f(x)

Vedclass · Accepted Answer

$\frac{12}{5}$

Vedclass · Answer

(B) The given differential equation is $(1+\cos^2 x) f'(x) - f(x) \sin 2x = 4 \sin 2x$. Divide by $(1+\cos^2 x)$ to get the linear form $f'(x) - f(x) \frac{\sin 2x}{1+\cos^2 x} = \frac{4 \sin 2x}{1+\cos^2 x}$. This is a linear differential equation of the form $f'(x) + P(x)f(x) = Q(x)$, where $P(x) = -\frac{\sin 2x}{1+\cos^2 x}$ and $Q(x) = \frac{4 \sin 2x}{1+\cos^2 x}$. The integrating factor $IF = e^{\int P(x) dx} = e^{\int -\frac{\sin 2x}{1+\cos^2 x} dx}$. Let $u = 1+\cos^2 x$, then $du = 2 \cos x (-\sin x) dx = -\sin 2x dx$. So, $IF = e^{\int \frac{du}{u}} = e^{\ln u} = 1+\cos^2 x$. The general solution is $f(x) \cdot IF = \int Q(x) \cdot IF dx + C$. $f(x)(1+\cos^2 x) = \int \frac{4 \sin 2x}{1+\cos^2 x} (1+\cos^2 x) dx = \int 4 \sin 2x dx = -2 \cos 2x + C$. Given $f(0)=0$, we have $0(1+1) = -2 \cos(0) + C \implies 0 = -2 + C \implies C = 2$. Thus, $f(x)(1+\cos^2 x) = 2 - 2 \cos 2x = 2(1 - \cos 2x) = 4 \sin^2 x$. $f(x) = \frac{4 \sin^2 x}{1+\cos^2 x}$. At $x = \frac{\pi}{3}$, $\sin^2(\frac{\pi}{3}) = (\frac{\sqrt{3}}{2})^2 = \frac{3}{4}$ and $\cos^2(\frac{\pi}{3}) = (\frac{1}{2})^2 = \frac{1}{4}$. $f(\frac{\pi}{3}) = \frac{4(3/4)}{1+1/4} = \frac{3}{5/4} = \frac{12}{5}$.

If $y=f(x)$ is the solution of the differential equation $(1+\cos^2 x) f'(x) - f(x) \sin 2x = 4 \sin 2x$ with $f(0)=0$, then $f(\frac{\pi}{3})=$

Similar Questions

If $\frac{dy}{dx} + y \tan x = \sin 2x$ and $y(0) = 1$,then $y(\pi)$ is equal to

The Integrating Factor $(I.F.)$ of the differential equation $\frac{dy}{dx} - \frac{3x^2y}{1+x^3} = \frac{\sin^2(x)}{1+x}$ is:

Let $f$ be a differentiable function with $\lim_{x \rightarrow \infty} f(x) = 0$. If $y^{\prime} + y f^{\prime}(x) - f(x) f^{\prime}(x) = 0$ and $\lim_{x \rightarrow \infty} y(x) = 0$, then (where $y^{\prime} = \frac{dy}{dx}$):

The solution of $y' - y = 1, y(0) = -1$ is given by $y(x) = $

The solution of $\frac{dx}{dy} + \frac{x}{y} = x^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module