Solve (1+y^2)dx = (tan^{-1}y - x)dy.

Steps to Solve

1. Find the integrating factor

The integrating factor (I.F.) is given by $e^{\int P dy}$. In this case, $P = \frac{1}{1+y^2}$, so we have: $$I.F. = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1}y}$$

2. Multiply the differential equation by the integrating factor

Multiply both sides of the equation $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan^{-1}y}{1+y^2}$ by $e^{\tan^{-1}y}$: $$e^{\tan^{-1}y}\frac{dx}{dy} + \frac{x}{1+y^2}e^{\tan^{-1}y} = \frac{\tan^{-1}y}{1+y^2}e^{\tan^{-1}y}$$

3. Recognize the left side as the derivative of a product

The left-hand side (LHS) is the derivative of $x \cdot e^{\tan^{-1}y}$ with respect to $y$: $$\frac{d}{dy}(x e^{\tan^{-1}y}) = e^{\tan^{-1}y}\frac{dx}{dy} + x \cdot e^{\tan^{-1}y} \cdot \frac{1}{1+y^2}$$ Which matches with the LHS from step 2. Thus, $$\frac{d}{dy}(x e^{\tan^{-1}y}) = \frac{\tan^{-1}y}{1+y^2}e^{\tan^{-1}y}$$

4. Integrate both sides with respect to $y$

Integrate both sides of the equation with respect to $y$: $$\int \frac{d}{dy}(x e^{\tan^{-1}y}) dy = \int \frac{\tan^{-1}y}{1+y^2}e^{\tan^{-1}y} dy$$ $$x e^{\tan^{-1}y} = \int \frac{\tan^{-1}y}{1+y^2}e^{\tan^{-1}y} dy$$

5. Solve the integral on the right-hand side

Let $u = \tan^{-1}y$. Then, $\frac{du}{dy} = \frac{1}{1+y^2}$, so $du = \frac{1}{1+y^2} dy$. The integral becomes: $$\int u e^u du$$ Using integration by parts, let $v = u$ and $dw = e^u du$. Then, $dv = du$ and $w = e^u$. $$\int u e^u du = u e^u - \int e^u du = u e^u - e^u + C$$ Substituting back $u = \tan^{-1}y$: $$\int \frac{\tan^{-1}y}{1+y^2}e^{\tan^{-1}y} dy = (\tan^{-1}y) e^{\tan^{-1}y} - e^{\tan^{-1}y} + C$$

6. Solve for $x$

Substitute the result of the integral back into the equation: $$x e^{\tan^{-1}y} = (\tan^{-1}y) e^{\tan^{-1}y} - e^{\tan^{-1}y} + C$$ Divide both sides by $e^{\tan^{-1}y}$: $$x = \tan^{-1}y - 1 + \frac{C}{e^{\tan^{-1}y}}$$ $$x = \tan^{-1}y - 1 + Ce^{-\tan^{-1}y}$$