Related Formula
Leibniz Integral Rule for differentiation:
ddx(∫₀^x f(t)dt) = f(x)$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_0^x f(t)\mathrm{d}t\right) = f(x)$$
Core Logic
Differentiate the given integral relation using Leibniz rule [cite: 1344]:
ddx[∫₀xg(t)dt] = ddx[x-∫₀xtg(t)dt]$$\frac{\mathrm{d}}{\mathrm{d}x}\left[\int_{0}^{x}g(t)dt\right] = \frac{\mathrm{d}}{\mathrm{d}x}\left[x-\int_{0}^{x}tg(t)dt\right]$$ [cite: 1344]
g(x) = 1 - xg(x) g(x)(1+x) = 1 g(x) = (1)/(1+x)$$g(x) = 1 - xg(x) \implies g(x)(1+x) = 1 \implies g(x) = \frac{1}{1+x}$$ [cite: 1345]
Substitute g(x)$g(x)$ into target differential equation configuration [cite: 1346]:
dydx - y x = 2(x+1) x · ((1)/(1+x)) = 2 x$$\frac{\mathrm{d}y}{\mathrm{d}x} - y\tan x = 2(x+1)\sec x \cdot \left(\frac{1}{1+x}\right) = 2\sec x$$ [cite: 1346]
Step 1: Finding the Integrating Factor
This matches a linear form dydx + P(x)y = Q(x)$\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)$ where P(x) = - x$P(x) = -\tan x$.
I.F. = e∫ - x dx = eln| x| = x$$\text{I.F.} = e^{\int -\tan x \, \mathrm{d}x} = e^{\ln|\cos x|} = \cos x$$ [cite: 1346]
Write general functional solution template [cite: 1348]:
y · x = ∫ (2 x · x) dx = ∫ 2 dx = 2x + C$$y \cdot \cos x = \int (2\sec x \cdot \cos x) \, \mathrm{d}x = \int 2 \, \mathrm{d}x = 2x + C$$ [cite: 1348]
Given boundary condition y(0) = 0 0 = 0 + C C = 0$y(0) = 0 \implies 0 = 0 + C \implies C = 0$ [cite: 1349].
y(x) = (2x)/( x) = 2x x$$y(x) = \frac{2x}{\cos x} = 2x\sec x$$ [cite: 1350]
Step 2: Numeric substitution
Substitute variable parameter values x = (π)/(3)$x = \frac{\pi}{3}$ [cite: 1352]:
y((π)/(3)) = 2((π)/(3)) ((π)/(3)) = (2π)/(3) · 2 = (4π)/(3)$$y\left(\frac{\pi}{3}\right) = 2\left(\frac{\pi}{3}\right)\sec\left(\frac{\pi}{3}\right) = \frac{2\pi}{3} \cdot 2 = \frac{4\pi}{3}$$ [cite: 1351]
Pattern Recognition
Integral functional definitions are codes for simpler underlying derivatives. Applying Leibniz rule immediately extracts the true variable functions.
Chapter Mix
Class 12 Mathematics: Differential Equations
