Related Formula
Integrating factor (I.F.) for a linear differential equation (dy)/(dx) + Py = Q$\frac{dy}{dx} + Py = Q$:
I.F. = e∫ P dx$$\text{I.F.} = e^{\int P \, dx}$$
General solution:
y · (I.F.) = ∫ Q · (I.F.) dx$$y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) \, dx$$
Core Logic
Given P = x$P = \tan x$, compute Integrating Factor:
I.F. = e∫ x dx = eln( x) = x$$\text{I.F.} = e^{\int \tan x \, dx} = e^{\ln(\sec x)} = \sec x$$
Set up integrated expression solution layout:
y · x = ∫ (2 + x)/((1 + 2 x)²) · x dx = ∫ (2 x + 1)/(( x + 2)²) · dx$$y \cdot \sec x = \int \frac{2 + \sec x}{(1 + 2\sec x)^2} \cdot \sec x \, dx = \int \frac{2\cos x + 1}{(\cos x + 2)^2} \cdot dx$$
Step 1: Evaluate Integration with Half-Angle Substitutions
Using tangent half-angle substitution t = (x)/(2)$t = \tan\frac{x}{2}$ transformations simplifies the integral loop structure down to:
y · x = (2)/(t + (3)/(t)) + C$$y \cdot \sec x = \frac{2}{t + \frac{3}{t}} + C$$
Plugging entry condition parameters f((π)/(3)) = √(3)10$f\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{10}$ tracking t = 1√(3)$t = \frac{1}{\sqrt{3}}$ explicitly isolates boundary condition constant C$C$:
C = 0$C = 0$
Step 2: Calculate Target Point Value
At target query point x = (π)/(4)$x = \frac{\pi}{4}$, half-angle parameters scale to t = √(2) - 1$t = \sqrt{2} - 1$:
y · √(2) = 2√(2) - 1 + 3√(2) - 1 = 2(√(2) - 1)6 - 2√(2)$$y \cdot \sqrt{2} = \frac{2}{\sqrt{2} - 1 + \frac{3}{\sqrt{2} - 1}} = \frac{2(\sqrt{2} - 1)}{6 - 2\sqrt{2}}$$
y = 4 - √(2)14$$y = \frac{4 - \sqrt{2}}{14}$$
Pattern Recognition
When integrating complex rational expressions involving trigonometric values, half-angle substitution methods (t = (x)/(2)$t = \tan\frac{x}{2}$) are standard for reducing polynomial degrees.
Chapter Mix
Class 12 Mathematics: Differential Equations