Related Formula
Standard first-order linear differential equation structure:
(dy)/(dx) + P(x)y = Q(x)$$\frac{dy}{dx} + P(x)y = Q(x)$$
Integrating Factor (I.F.) = e∫ P(x)dx$$\text{Integrating Factor (I.F.)} = e^{\int P(x)dx}$$
Step 1: Reduce into Standard Format
Divide the full expression by 2 x$2\cos x$:
(dy)/(dx) = (2 x x)/(2 x) - (4y x)/(2 x)$$\frac{dy}{dx} = \frac{2\sin x \cos x}{2\cos x} - \frac{4y\sin x}{2\cos x}$$
(dy)/(dx) + 2y x = x$$\frac{dy}{dx} + 2y\tan x = \sin x$$
Step 2: Integrating Factor & Solution
Compute the integrating multiplier :
I.F. = e∫ 2 x dx = e2ln| x| = ² x$$\text{I.F.} = e^{\int 2\tan x dx} = e^{2\ln|\sec x|} = \sec^2 x$$
Write general integration solution path :
y · ² x = ∫ x · ² x dx = ∫ x x dx = x + C$$y \cdot \sec^2 x = \int \sin x \cdot \sec^2 x dx = \int \sec x \tan x dx = \sec x + C$$
y = x + C ² x$$y = \cos x + C\cos^2 x$$
Step 3: Boundary Evaluation
Apply the initialization condition y((π)/(3)) = 0$y\left(\frac{\pi}{3}\right) = 0$:
0 = ((π)/(3)) + C ²((π)/(3)) ⇒ 0 = (1)/(2) + C((1)/(4)) ⇒ C = -2$$0 = \cos\left(\frac{\pi}{3}\right) + C\cos^2\left(\frac{\pi}{3}\right) \Rightarrow 0 = \frac{1}{2} + C\left(\frac{1}{4}\right) \Rightarrow C = -2$$
Thus, the solution is y = x - 2 ² x$y = \cos x - 2\cos^2 x$ .
Find derivative y$y\prime$ :
y = - x + 4 x x = - x + 2 2x$$y\prime = -\sin x + 4\cos x \sin x = -\sin x + 2\sin 2x$$
Step 4: Target Calculation
Evaluate components at x = (π)/(4)$x = \frac{\pi}{4}$ :
y((π)/(4)) = 1√(2) - 2((1)/(2)) = 1√(2) - 1$$y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} - 2\left(\frac{1}{2}\right) = \frac{1}{\sqrt{2}} - 1$$
y ((π)/(4)) = - 1√(2) + 2 ((π)/(2)) = - 1√(2) + 2$$y\prime\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} + 2\sin\left(\frac{\pi}{2}\right) = -\frac{1}{\sqrt{2}} + 2$$
y ((π)/(4)) + y((π)/(4)) = (- 1√(2) + 2) + ( 1√(2) - 1) = 1$$y\prime\left(\frac{\pi}{4}\right) + y\left(\frac{\pi}{4}\right) = \left(-\frac{1}{\sqrt{2}} + 2\right) + \left(\frac{1}{\sqrt{2}} - 1\right) = 1$$
Pattern Recognition
Linear standard layout conversions depend entirely on clear integrating factor reductions. Remember ∫ x dx = ln| x|$\int \tan x dx = \ln|\sec x|$ clearly to safely output exact matching polynomial definitions.
Chapter Mix
Class 12 Mathematics: Differential Equations
