Related Formula
Standard Form of LDE: (dy)/(dx) + P(x)y = Q(x)$$\text{Standard Form of LDE: } \frac{dy}{dx} + P(x)y = Q(x)$$
I.F. = e∫ P(x)dx$$I.F. = e^{\int P(x)dx}$$
Solution: y · (I.F.) = ∫ Q(x) · (I.F.) dx + C$$\text{Solution: } y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$$
Core Logic
Given the differential equation:
y 2x + x - (1+ ² x)(dy)/(dx) = 0$$y\sin 2x + \sin x - (1+\cos^2 x)\frac{dy}{dx} = 0$$
Rearrange it into the standard linear differential equation format:
(1+ ² x)(dy)/(dx) - y 2x = x$$(1+\cos^2 x)\frac{dy}{dx} - y\sin 2x = \sin x$$
(dy)/(dx) - (( 2x)/(1+ ² x))y = ( x)/(1+ ² x)$$\frac{dy}{dx} - \left(\frac{\sin 2x}{1+\cos^2 x}\right)y = \frac{\sin x}{1+\cos^2 x}$$
Step 1: Find the Integrating Factor (I.F.)
Here, P(x) = -( 2x)/(1+ ² x)$P(x) = -\frac{\sin 2x}{1+\cos^2 x}$.
I.F. = e∫ -( 2x)/(1+ ² x) dx$$I.F. = e^{\int -\frac{\sin 2x}{1+\cos^2 x} dx}$$
Let 1+ ² x = t$1+\cos^2 x = t$, then dt = -2 x x dx = - 2x dx$dt = -2\cos x\sin x \, dx = -\sin 2x \, dx$.
I.F. = e∫ (1)/(t) dt = eln t = t$$I.F. = e^{\int \frac{1}{t} dt} = e^{\ln t} = t$$
I.F. = 1+ ² x$$I.F. = 1+\cos^2 x$$
Step 2: Solve the Differential Equation
Multiply the LDE by the integrating factor:
y · (1+ ² x) = ∫ (( x)/(1+ ² x)) · (1+ ² x) dx$$y \cdot (1+\cos^2 x) = \int \left(\frac{\sin x}{1+\cos^2 x}\right) \cdot (1+\cos^2 x) dx$$
y(1+ ² x) = ∫ x dx$$y(1+\cos^2 x) = \int \sin x dx$$
y(1+ ² x) = - x + C$$y(1+\cos^2 x) = -\cos x + C$$
Apply the initial condition f(0) = 0$f(0) = 0$ (which means y=0$y=0$ when x=0$x=0$):
0 · (1+1) = -1 + C ⇒ C = 1$$0 \cdot (1+1) = -1 + C \Rightarrow C = 1$$
The particular solution is:
y(1+ ² x) = 1 - x$$y(1+\cos^2 x) = 1 - \cos x$$
Step 3: Evaluate at x = pi/2
To find f((π)/(2))$f(\frac{\pi}{2})$, substitute x = (π)/(2)$x = \frac{\pi}{2}$:
y(1+ ²((π)/(2))) = 1 - ((π)/(2))$$y\left(1+\cos^2\left(\frac{\pi}{2}\right)\right) = 1 - \cos\left(\frac{\pi}{2}\right)$$
y(1+0) = 1 - 0$y(1+0) = 1 - 0$
y = 1$y = 1$
Pattern Recognition
Whenever y$y$ and y'$y'$ appear linearly separated by complex trigonometric polynomials, format aggressively into standard LDE form. The derivative of 1+ ²x$1+\cos^2x$ is exactly - 2x$-\sin 2x$, making the logarithmic integration seamless.
Chapter Mix
Class 12 Mathematics: Differential Equations
Class 12 Mathematics: Integral Calculus
