Solve \( \left(1+y^{2}\right) d x=\left(\operatorname{Tan}^{-1} y-x\right) d y \)

We start with the differential equation \[ \left(1+y^2\right)dx = \left(\tan^{-1}y - x\right)dy. \] ### Step 1. Write the Equation in Standard Linear Form Divide both sides by \(1+y^2\) (assuming \(1+y^2 \neq 0\)) to obtain \[ \frac{dx}{dy} = \frac{\tan^{-1} y - x}{1+y^2}. \] Rearrange the terms to cast it into the standard linear form \[ \frac{dx}{dy} + \frac{1}{1+y^2} \, x = \frac{\tan^{-1} y}{1+y^2}. \] ### Step 2. Find the Integrating Factor The integrating factor \(\mu(y)\) is given by \[ \mu(y) = \exp\left(\int \frac{1}{1+y^2}\,dy\right). \] The integral is \[ \int \frac{1}{1+y^2}\,dy = \tan^{-1} y, \] so the integrating factor is \[ \mu(y) = e^{\tan^{-1} y}. \] ### Step 3. Multiply the Equation by the Integrating Factor Multiply every term in the differential equation by \(e^{\tan^{-1} y}\): \[ e^{\tan^{-1} y}\frac{dx}{dy} + \frac{e^{\tan^{-1} y}}{1+y^2}\, x = \frac{\tan^{-1} y}{1+y^2}\,e^{\tan^{-1} y}. \] Notice that the left-hand side is the derivative of \(x\,e^{\tan^{-1} y}\): \[ \frac{d}{dy}\left(x\,e^{\tan^{-1} y}\right) = \frac{\tan^{-1} y}{1+y^2}e^{\tan^{-1} y}. \] ### Step 4. Integrate Both Sides Integrate both sides with respect to \(y\): \[ \int \frac{d}{dy}\left(x\,e^{\tan^{-1} y}\right) \, dy = \int \frac{\tan^{-1} y}{1+y^2}e^{\tan^{-1} y}\,dy. \] The left-hand side integrates to \[ x\,e^{\tan^{-1} y}. \] For the right-hand side, use the substitution \[ u = \tan^{-1} y \quad \text{so that} \quad \frac{du}{dy} = \frac{1}{1+y^2} \quad \Longrightarrow \quad dy = (1+y^2)\,du. \] Then the integral becomes \[ \int u\,e^u\,du. \] ### Step 5. Evaluate the Integral \(\int u\,e^u\,du\) Use integration by parts: Let \[ v = u \quad \text{and} \quad dw = e^u\,du. \] Then \[ dv = du \quad \text{and} \quad w = e^u. \] Thus, \[ \int u\,e^u\,du = u\,e^u - \int e^u\,du = u\,e^u - e^u + C = e^u(u-1) + C. \] Returning to the variable \(y\) (\(u=\tan^{-1} y\)), we have \[ \int \frac{\tan^{-1} y}{1+y^2}e^{\tan^{-1} y}\,dy = e^{\tan^{-1} y}\left(\tan^{-1} y - 1\right) + C. \] ### Step 6. Write the General Solution We have \[ x\,e^{\tan^{-1} y} = e^{\tan^{-1} y}\left(\tan^{-1} y - 1\right) + C. \] Divide through by \(e^{\tan^{-1} y}\) (which is never zero) to obtain the general solution: \[ x = \tan^{-1} y - 1 + C\,e^{-\tan^{-1} y}. \] This is the solution to the given differential equation.