Question

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

Hint:

If a first-order differential equation doesn't seem to be separable or linear in the form \(\frac{dy}{dx}\), always check if it becomes linear by treating \(x\) as the function of \(y\), i.e., by rearranging it into the form \(\frac{dx}{dy} + P(y)x = Q(y)\). This is a common pattern in exam questions.

Answer 1

Step 1: Understanding the Concept:
The given equation is a first-order differential equation. It is not easily separable. We can try to rearrange it into the form of a linear differential equation. By rewriting it with \(x\) as the dependent variable and \(y\) as the independent variable, we can see that it fits the standard linear form \(\frac{dx}{dy} + P(y)x = Q(y)\).
Step 2: Key Formula or Approach:
1. Rearrange the equation into the standard linear form.
2. Identify the functions \(P(y)\) and \(Q(y)\).
3. Calculate the integrating factor (I.F.) using the formula I.F. = \(e^{\int P(y) dy}\).
4. The general solution is given by the formula: \(x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) dy + C\).
5. Solve the resulting integral, which may require techniques like integration by parts.
Step 3: Detailed Explanation:
The given differential equation is: \[ (\tan^{-1}y - x)dy = (1+y^2)dx \] Rearrange to get \(\frac{dx}{dy}\): \[ \frac{dx}{dy} = \frac{\tan^{-1}y - x}{1+y^2} \] \[ \frac{dx}{dy} = \frac{\tan^{-1}y}{1+y^2} - \frac{x}{1+y^2} \] Bring the term with \(x\) to the LHS to match the standard linear form: \[ \frac{dx}{dy} + \frac{1}{1+y^2}x = \frac{\tan^{-1}y}{1+y^2} \] This is a linear differential equation in \(x\).
1. Identify P(y) and Q(y): \[ P(y) = \frac{1}{1+y^2}, \quad Q(y) = \frac{\tan^{-1}y}{1+y^2} \] 2. Calculate the Integrating Factor (I.F.): \[ \text{I.F.} = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1}y} \] 3. Apply the solution formula: \[ x \cdot e^{\tan^{-1}y} = \int \left(\frac{\tan^{-1}y}{1+y^2}\right) e^{\tan^{-1}y} dy + C \] To solve the integral on the RHS, let \(t = \tan^{-1}y\). Then \(dt = \frac{1}{1+y^2} dy\). The integral becomes: \[ \int t e^t dt \] We solve this using integration by parts, \(\int u dv = uv - \int v du\).
Let \(u = t\) and \(dv = e^t dt\). Then \(du = dt\) and \(v = e^t\). \[ \int t e^t dt = t e^t - \int e^t dt = t e^t - e^t = e^t(t-1) \] Substituting back \(t = \tan^{-1}y\), the integral is \(e^{\tan^{-1}y}(\tan^{-1}y - 1)\).
The general solution is: \[ x \cdot e^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y - 1) + C \] 4. Isolate x: Divide the entire equation by \(e^{\tan^{-1}y}\): \[ x = (\tan^{-1}y - 1) + C e^{-\tan^{-1}y} \] Step 4: Final Answer:
The solution to the differential equation is \(x = \tan^{-1}y - 1 + C e^{-\tan^{-1}y}\), where C is an arbitrary constant.