Let $g$ be a differentiable function such that $\int_0^x g(t)dt=x-\int_0^x tg(t)dt$, $x\ge0$ and let $y = y(x)$ satisfy the differential equation $\frac{dy}{dx} - y\tan x = 2(x+1)\sec x g(x)$, $x\in [0,\frac{\pi}{2})$. If $y(0) = 0$, then $y(\frac{\pi}{3})$ is equal to
$\frac{4\pi}{3}$
Finding the Differentiable Function g(x)
We begin by differentiating the given integral equation with respect to $x$: $ \int_0^x g(t)dt = x - \int_0^x tg(t)dt $ Applying the Fundamental Theorem of Calculus to both sides: $ \frac{d}{dx}\left(\int_0^x g(t)dt\right) = \frac{d}{dx}(x) - \frac{d}{dx}\left(\int_0^x tg(t)dt\right) $ This yields: $ g(x) = 1 - xg(x) $ To find $g(x)$, we rearrange the equation: $ g(x) + xg(x) = 1 $ $ g(x)(1+x) = 1 $ $ g(x) = \frac{1}{1+x} $
Solving the Differential Equation
The given first-order linear differential equation is: $ \frac{dy}{dx} - y\tan x = 2(x+1)\sec x g(x) $ Substitute the found expression for $g(x)$: $ \frac{dy}{dx} - y\tan x = 2(x+1)\sec x \left(\frac{1}{1+x}\right) $ This simplifies to: $ \frac{dy}{dx} - y\tan x = 2\sec x $ This is in the form $\frac{dy}{dx} + P(x)y = Q(x)$, with $P(x) = -\tan x$ and $Q(x) = 2\sec x$. The integrating factor (IF) is $e^{\int P(x)dx}$: $ IF = e^{\int -\tan x dx} = e^{\ln|\cos x|} $ Since $x \in [0, \frac{\pi}{2})$, $\cos x > 0$, so $IF = \cos x$. Multiplying the differential equation by the IF: $ (\cos x)\frac{dy}{dx} - y(\cos x)\tan x = 2\sec x \cos x $ The left side is the derivative of $(y \cdot IF)$: $ \frac{d}{dx}(y \cos x) = 2 $ Integrating both sides: $ y \cos x = \int 2 dx = 2x + C $ The general solution is: $ y(x) = \frac{2x + C}{\cos x} = (2x+C)\sec x $
Applying the Initial Condition y(0)=0
Use the initial condition $y(0) = 0$ to find the constant $C$. Substitute $x=0$ into the general solution: $ y(0) = (2(0) + C)\sec(0) $ $ 0 = (C)(1) $ $ C = 0 $ The particular solution is therefore: $ y(x) = 2x\sec x $
Calculating y(pi/3)
To find $y(\frac{\pi}{3})$, substitute $x = \frac{\pi}{3}$ into the particular solution: $ y\left(\frac{\pi}{3}\right) = 2\left(\frac{\pi}{3}\right)\sec\left(\frac{\pi}{3}\right) $ We know that $\sec(\frac{\pi}{3}) = 2$. $ y\left(\frac{\pi}{3}\right) = 2\left(\frac{\pi}{3}\right)(2) $ $ y\left(\frac{\pi}{3}\right) = \frac{4\pi}{3} $
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