Let $y = y(x)$ be the solution curve of the differential equation $\left( 1 + x^{2} \right)dy + \lef

Question

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

Accepted Answer

$\frac{dy}{dx} + \frac{y}{x^{2} + 1} = \frac{\tan^{- 1}x}{x^{2} + 1}$

$$\begin{matrix} & \text{~}\text{I.f.}\text{~} = e^{\tan^{- 1}x} \\ & y \times e^{\tan^{- 1}x} = \int_{}^{}\ e^{\tan^{- 1}x} \cdot \frac{\tan^{- 1}x}{1 + x^{2}}dx \end{matrix}$$

$$\begin{matrix} & y(0) = 1 \Rightarrow c = 2 \\ & y(1) = \frac{2}{e^{\pi/4}} + \frac{\pi}{4} - 1 \end{matrix}$$

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