Solve the differential equation: $(1+y^2)dx = (\tan^{-1}y - x)dy$

Steps to Solve

1. Rewrite the given differential equation

The given differential equation is $(1+y^2)dx = (\tan^{-1}y - x)dy$. We rewrite it to the form $\frac{dx}{dy} = \frac{\tan^{-1}y - x}{1+y^2}$.

2. Separate the terms

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

3. Rearrange the equation into a linear first-order form

Rearrange the terms so that it looks like a linear first-order differential equation $\frac{dx}{dy} + P(y)x = Q(y)$. Thus, $\frac{dx}{dy} + \frac{1}{1+y^2}x = \frac{\tan^{-1}y}{1+y^2}$.

4. Find the integrating factor

The integrating factor is given by $e^{\int P(y)dy}$. In our equation, $P(y) = \frac{1}{1+y^2}$. Then, $\int P(y)dy = \int \frac{1}{1+y^2}dy = \tan^{-1}y$. Hence, the integrating factor is $e^{\tan^{-1}y}$.

5. Multiply the entire equation by the integrating factor

Multiply the equation $\frac{dx}{dy} + \frac{1}{1+y^2}x = \frac{\tan^{-1}y}{1+y^2}$ by $e^{\tan^{-1}y}$. We have: $e^{\tan^{-1}y}\frac{dx}{dy} + e^{\tan^{-1}y}\frac{1}{1+y^2}x = e^{\tan^{-1}y}\frac{\tan^{-1}y}{1+y^2}$.

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

The left side is the derivative of $x \cdot e^{\tan^{-1}y}$ with respect to $y$. Thus we have $\frac{d}{dy}(x e^{\tan^{-1}y}) = e^{\tan^{-1}y}\frac{\tan^{-1}y}{1+y^2}$.

7. Integrate both sides with respect to y

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

8. Solve the integral

Let $u = \tan^{-1}y$, then $du = \frac{1}{1+y^2}dy$. The integral becomes $\int ue^u du$. Using integration by parts: $\int ue^u du = u\int e^u du - \int (\frac{du}{du}\int e^u du) du = ue^u - \int e^u du = ue^u - e^u + C$. Substituting back $u = \tan^{-1}y$, we have $\int e^{\tan^{-1}y}\frac{\tan^{-1}y}{1+y^2} dy = (\tan^{-1}y)e^{\tan^{-1}y} - e^{\tan^{-1}y} + C$.

9. Substitute back into the solution

Now we have $x e^{\tan^{-1}y} = (\tan^{-1}y)e^{\tan^{-1}y} - e^{\tan^{-1}y} + C$.

10. Solve for x

Divide by $e^{\tan^{-1}y}$: $x = \tan^{-1}y - 1 + Ce^{-\tan^{-1}y}$.