Related Formula
For a linear differential equation of the form (dy)/(dx) + P(x)y = Q(x)$\frac{dy}{dx} + P(x)y = Q(x)$:
Integrating Factor (IF) = e∫ P(x)dx$$\text{Integrating Factor (IF)} = e^{\int P(x)dx}$$
Solution is y · IF = ∫ (Q(x) · IF) dx + C$$\text{Solution is } y \cdot \text{IF} = \int (Q(x) \cdot \text{IF}) dx + C$$
Core Logic
Rearrange the given differential equation to standard linear form:
(1 + x²)dy = ( ⁻¹x - y)dx$(1 + x^2)dy = (\tan^{-1}x - y)dx$
(dy)/(dx) + (y)/(x² + 1) = ⁻¹xx² + 1$$\frac{dy}{dx} + \frac{y}{x^2 + 1} = \frac{\tan^{-1}x}{x^2 + 1}$$
Step 1: Find the Integrating Factor
P(x) = (1)/(x² + 1)$$P(x) = \frac{1}{x^2 + 1}$$
IF = e∫ (1)/(x² + 1) dx = e^ ⁻¹x$$\text{IF} = e^{\int \frac{1}{x^2 + 1} dx} = e^{\tan^{-1}x}$$
Step 2: Solve the Integral
y · e^ ⁻¹x = ∫ e^ ⁻¹x · ⁻¹x1 + x² dx$$y \cdot e^{\tan^{-1}x} = \int e^{\tan^{-1}x} \cdot \frac{\tan^{-1}x}{1 + x^2} dx$$
Let t = ⁻¹x$t = \tan^{-1}x$, then dt = (1)/(1 + x²) dx$dt = \frac{1}{1 + x^2} dx$.
The integral becomes ∫ t e^t dt$\int t e^t dt$.
Using integration by parts:
∫ t e^t dt = t e^t - e^t + C$$\int t e^t dt = t e^t - e^t + C$$
Substituting back t = ⁻¹x$t = \tan^{-1}x$:
y · e^ ⁻¹x = ⁻¹x(e^ ⁻¹x) - e^ ⁻¹x + C$$y \cdot e^{\tan^{-1}x} = \tan^{-1}x(e^{\tan^{-1}x}) - e^{\tan^{-1}x} + C$$
Step 3: Apply Boundary Condition
Given y(0) = 1$y(0) = 1$:
1 · e⁰ = 0 · e⁰ - e⁰ + C$$1 \cdot e^0 = 0 \cdot e^0 - e^0 + C$$
1 = 0 - 1 + C ⇒ C = 2$$1 = 0 - 1 + C \Rightarrow C = 2$$
Step 4: Evaluate at x = 1
Equation of curve:
y = ⁻¹x - 1 + 2e^- ⁻¹x$$y = \tan^{-1}x - 1 + 2e^{-\tan^{-1}x}$$
Evaluate at x = 1$x = 1$:
y(1) = ⁻¹(1) - 1 + 2e^- ⁻¹(1)$$y(1) = \tan^{-1}(1) - 1 + 2e^{-\tan^{-1}(1)}$$
y(1) = (π)/(4) - 1 + 2e-π/4$$y(1) = \frac{\pi}{4} - 1 + 2e^{-\pi/4}$$
y(1) = 2eπ/4 + (π)/(4) - 1$$y(1) = \frac{2}{e^{\pi/4}} + \frac{\pi}{4} - 1$$
Pattern Recognition
A classic LDE integration trap: ∫ ef(x) f(x) f'(x) dx$\int e^{f(x)} f(x) f'(x) dx$ resolves trivially with the substitution u = f(x)$u = f(x)$ turning it into ∫ u e^u du$\int u e^u du$, which always evaluates to e^u(u - 1) + C$e^u(u - 1) + C$.
Chapter Mix
Class 12 Maths: Differential Equations
Class 12 Maths: Integrals
