Let y = y(x)$y = y(x)$ be the solution curve of the differential equation (1 + x^2)dy + (y - tan^-1x)dx = 0$(1 + x^2)dy + (y - \tan^{-1}x)dx = 0$ , y(0) = 1$y(0) = 1$ . Then the value of y(1)$y(1)$ is:

### Related Formula For a linear differential equation of the form fracdydx + P(x)y = Q(x)$\frac{dy}{dx} + P(x)y = Q(x)$: textIntegrating Factor (IF) = e^int P(x)dx$$\text{Integrating Factor (IF)} = e^{\int P(x)dx}$$ textSolution is y cdot textIF = int (Q(x) cdot textIF) 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 + x^2)dy = (tan^-1x - y)dx$(1 + x^2)dy = (\tan^{-1}x - y)dx$ fracdydx + fracyx^2 + 1 = fractan^-1xx^2 + 1$$\frac{dy}{dx} + \frac{y}{x^2 + 1} = \frac{\tan^{-1}x}{x^2 + 1}$$ ### Step 1: Find the Integrating Factor P(x) = frac1x^2 + 1$$P(x) = \frac{1}{x^2 + 1}$$ textIF = e^int frac1x^2 + 1 dx = e^tan^-1x$$\text{IF} = e^{\int \frac{1}{x^2 + 1} dx} = e^{\tan^{-1}x}$$ ### Step 2: Solve the Integral y cdot e^tan^-1x = int e^tan^-1x cdot fractan^-1x1 + x^2 dx$$y \cdot e^{\tan^{-1}x} = \int e^{\tan^{-1}x} \cdot \frac{\tan^{-1}x}{1 + x^2} dx$$ Let t = tan^-1x$t = \tan^{-1}x$, then dt = frac11 + x^2 dx$dt = \frac{1}{1 + x^2} dx$. The integral becomes int t e^t dt$\int t e^t dt$. Using integration by parts: int t e^t dt = t e^t - e^t + C$$\int t e^t dt = t e^t - e^t + C$$ Substituting back t = tan^-1x$t = \tan^{-1}x$: y cdot e^tan^-1x = tan^-1x(e^tan^-1x) - e^tan^-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 cdot e^0 = 0 cdot e^0 - e^0 + C$$1 \cdot e^0 = 0 \cdot e^0 - e^0 + C$$ 1 = 0 - 1 + C Rightarrow C = 2$$1 = 0 - 1 + C \Rightarrow C = 2$$ ### Step 4: Evaluate at x = 1 Equation of curve: y = tan^-1x - 1 + 2e^-tan^-1x$$y = \tan^{-1}x - 1 + 2e^{-\tan^{-1}x}$$ Evaluate at x = 1$x = 1$: y(1) = tan^-1(1) - 1 + 2e^-tan^-1(1)$$y(1) = \tan^{-1}(1) - 1 + 2e^{-\tan^{-1}(1)}$$ y(1) = fracpi4 - 1 + 2e^-pi/4$$y(1) = \frac{\pi}{4} - 1 + 2e^{-\pi/4}$$ y(1) = frac2e^pi/4 + fracpi4 - 1$$y(1) = \frac{2}{e^{\pi/4}} + \frac{\pi}{4} - 1$$ ### Pattern Recognition A classic LDE integration trap: int e^f(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 int u e^u du$\int u e^u du$, which always evaluates to e^u(u - 1) + C$e^u(u - 1) + C$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations Class 12 Maths: Integrals