Question

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

Hint:

If a first-order differential equation looks complicated and is not separable or homogeneous, always try rearranging it into the linear form \(\frac{dy}{dx} + P(x)y = Q(x)\). If that doesn't work, try rearranging it into the alternative linear form \(\frac{dx}{dy} + P(y)x = Q(y)\). One of these two forms often provides a straightforward path to the solution.

Answer 1

Step 1: Understanding the Concept:
This differential equation is not easily solvable in the form \(\frac{dy}{dx} = f(x,y)\). However, if we rearrange it to express \(\frac{dx}{dy}\), it takes the form of a linear differential equation in x.
Step 2: Key Formula or Approach:
The standard form of a first-order linear differential equation in x is: \[ \frac{dx}{dy} + P(y)x = Q(y) \] The solution is given by: \[ x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) dy + C \] where the integrating factor (I.F.) is \(e^{\int P(y) dy}\).
Step 3: Detailed Explanation:
The given equation is \((1+y^2) dx = (\tan^{-1} y - x) dy\).
Rearrange it to find \(\frac{dx}{dy}\): \[ \frac{dx}{dy} = \frac{\tan^{-1} y - x}{1+y^2} \] \[ \frac{dx}{dy} = \frac{\tan^{-1} y}{1+y^2} - \frac{x}{1+y^2} \] Bring the term with x to the left side to match the standard form: \[ \frac{dx}{dy} + \frac{1}{1+y^2}x = \frac{\tan^{-1} y}{1+y^2} \] This is a linear DE with \(P(y) = \frac{1}{1+y^2}\) and \(Q(y) = \frac{\tan^{-1} y}{1+y^2}\).
First, find the integrating factor (I.F.): \[ \text{I.F.} = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y} \] Now, apply the solution formula: \[ x \cdot e^{\tan^{-1} y} = \int \frac{\tan^{-1} y}{1+y^2} \cdot e^{\tan^{-1} y} dy + C \] To solve the integral on the right, let \(t = \tan^{-1} y\), so \(dt = \frac{1}{1+y^2} dy\). The integral becomes: \[ \int t e^t dt \] Using integration by parts (\(\int u dv = uv - \int v du\)) with \(u=t\) and \(dv=e^t dt\): \[ \int t e^t dt = t e^t - \int e^t dt = t e^t - e^t = e^t(t-1) \] Substitute back \(t = \tan^{-1} y\): \[ \int \frac{\tan^{-1} y}{1+y^2} e^{\tan^{-1} y} dy = e^{\tan^{-1} y}(\tan^{-1} y - 1) \] Substitute this back into the solution: \[ x \cdot e^{\tan^{-1} y} = e^{\tan^{-1} y}(\tan^{-1} y - 1) + C \] To get the explicit solution for x, divide the entire equation by \(e^{\tan^{-1} y}\): \[ x = \tan^{-1} y - 1 + \frac{C}{e^{\tan^{-1} y}} \] \[ x = \tan^{-1} y - 1 + Ce^{-\tan^{-1}y} \] Step 4: Final Answer:
The general solution of the differential equation is \(x = \tan^{-1} y - 1 + Ce^{-\tan^{-1}y}\).