The given differential equation is:
$(1 + x^2) dy + (y - \tan^{-1}x) dx = 0$
With the initial condition $y(0) = 1$. We need to find the value of $y(1)$.
Differential Equation Rearrangement
First, rearrange the equation into the standard linear first-order form $\frac{dy}{dx} + P(x)y = Q(x)$.
- Divide by $(1 + x^2)$: $dy + \frac{y - \tan^{-1}x}{1 + x^2} dx = 0$
- Isolate the $dy$ term: $dy = - \frac{y - \tan^{-1}x}{1 + x^2} dx$
- Rewrite as a derivative: $\frac{dy}{dx} = - \frac{y}{1 + x^2} + \frac{\tan^{-1}x}{1 + x^2}$
- Standard form: $\frac{dy}{dx} + \frac{1}{1 + x^2}y = \frac{\tan^{-1}x}{1 + x^2}$
Here, $P(x) = \frac{1}{1 + x^2}$ and $Q(x) = \frac{\tan^{-1}x}{1 + x^2}$.
Integrating Factor Calculation
Calculate the integrating factor (IF):
- $IF = e^{\int P(x) dx}$
- $\int P(x) dx = \int \frac{1}{1 + x^2} dx = \tan^{-1}x$
- $IF = e^{\tan^{-1}x}$
General Solution Derivation
The general solution is given by $y \cdot IF = \int Q(x) \cdot IF dx + C$.
- $y \cdot e^{\tan^{-1}x} = \int \frac{\tan^{-1}x}{1 + x^2} \cdot e^{\tan^{-1}x} dx + C$
- To solve the integral $\int \frac{\tan^{-1}x}{1 + x^2} e^{\tan^{-1}x} dx$, let $u = \tan^{-1}x$. Then $du = \frac{1}{1 + x^2} dx$.
- The integral becomes $\int u e^u du$.
- Using integration by parts ($\int w dv = wv - \int v dw$), let $w=u$ and $dv=e^u du$. Then $dw=du$ and $v=e^u$.
- $\int u e^u du = u e^u - \int e^u du = u e^u - e^u = (u-1)e^u$.
- Substitute back $u = \tan^{-1}x$: The integral is $(\tan^{-1}x - 1)e^{\tan^{-1}x}$.
- So, $y \cdot e^{\tan^{-1}x} = (\tan^{-1}x - 1)e^{\tan^{-1}x} + C$.
- Divide by $e^{\tan^{-1}x}$ to get the general solution for $y(x)$: $y(x) = (\tan^{-1}x - 1) + C e^{-\tan^{-1}x}$
Applying Initial Condition
Use the initial condition $y(0) = 1$ to find the constant $C$.
- $1 = (\tan^{-1}0 - 1) + C e^{-\tan^{-1}0}$
- $1 = (0 - 1) + C e^0$
- $1 = -1 + C \cdot 1$
- $C = 2$
The specific solution is $y(x) = (\tan^{-1}x - 1) + 2 e^{-\tan^{-1}x}$.
Calculating y(1)
Now, find the value of $y(1)$ using the specific solution.
- $y(1) = (\tan^{-1}1 - 1) + 2 e^{-\tan^{-1}1}$
- Since $\tan^{-1}1 = \frac{\pi}{4}$: $y(1) = (\frac{\pi}{4} - 1) + 2 e^{-\pi/4}$
- $y(1) = \frac{\pi}{4} - 1 + \frac{2}{e^{\pi/4}}$
- Rearranging the terms gives: $y(1) = \frac{2}{e^{\pi/4}} + \frac{\pi}{4} - 1$
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