Question

The solution of the differential equation,
\((x^2 + 1)\frac{dy}{dx} + 2xy = \sqrt{x^2 + 4}\), is

• A. \(y=(x^2+1)^{-1}\left(\frac{1}{2}x\sqrt{x^2+4}+2\log(x + \sqrt{x^2+4})\right)+c\); where c is a constant
• B. \(y=(x^2+1)^{-\frac{1}{2}}\left(\frac{1}{2}x\sqrt{x^2+4}+2\log(x + \sqrt{x^2+4})\right)+c\); where c is a constant
• C. \(y=\frac{1}{2}\left(\frac{1}{2}x\sqrt{x^2+4}+2\log|x + \sqrt{x^2+4}|\right)+c\); where c is a constant
• D. \(y=\left(x\sqrt{x^2+4}+2\log(x + \sqrt{x^2+4})\right)+c\); where c is a constant

Hint:

Always try to bring a first-order differential equation to the standard linear form \(\frac{dy}{dx} + P(x)y = Q(x)\). Recognizing this form is the key to applying the integrating factor method correctly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The given differential equation is a first-order linear differential equation. An equation of the form \(\frac{dy}{dx} + P(x)y = Q(x)\) can be solved using the integrating factor (I.F.) method.

Step 2: Key Formula or Approach:
1. Rewrite the equation in the standard form: \(\frac{dy}{dx} + P(x)y = Q(x)\).
2. Find the integrating factor: \(I.F. = e^{\int P(x)dx}\).
3. The solution is given by: \(y . (I.F.) = \int Q(x) . (I.F.) dx + C\).

Step 3: Detailed Explanation:
The given equation is \((x^2 + 1)\frac{dy}{dx} + 2xy = \sqrt{x^2 + 4}\).
First, we write it in the standard form by dividing by \((x^2+1)\): \[ \frac{dy}{dx} + \frac{2x}{x^2+1}y = \frac{\sqrt{x^2+4}}{x^2+1} \] Here, \(P(x) = \frac{2x}{x^2+1}\) and \(Q(x) = \frac{\sqrt{x^2+4}}{x^2+1}\).
Next, we find the integrating factor: \[ I.F. = e^{\int \frac{2x}{x^2+1}dx} \] Let \(u = x^2+1\), so \(du = 2xdx\). The integral becomes \(\int \frac{du}{u} = \ln(u) = \ln(x^2+1)\). \[ I.F. = e^{\ln(x^2+1)} = x^2+1 \] Now, the solution is: \[ y . (x^2+1) = \int \frac{\sqrt{x^2+4}}{x^2+1} . (x^2+1) dx + C \] \[ y(x^2+1) = \int \sqrt{x^2+4} dx + C \] We use the standard integration formula \(\int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\ln|x+\sqrt{x^2+a^2}|\).
With \(a^2=4\), we get: \[ \int \sqrt{x^2+4} dx = \frac{x}{2}\sqrt{x^2+4} + \frac{4}{2}\ln|x+\sqrt{x^2+4}| = \frac{1}{2}x\sqrt{x^2+4} + 2\ln(x+\sqrt{x^2+4}) \] So, the solution is: \[ y(x^2+1) = \frac{1}{2}x\sqrt{x^2+4} + 2\log(x+\sqrt{x^2+4}) + C \] Isolating \(y\), we get: \[ y = \frac{1}{x^2+1}\left(\frac{1}{2}x\sqrt{x^2+4} + 2\log(x+\sqrt{x^2+4}) + C\right) \]
Step 4: Final Answer:
The calculated solution matches the form of option (A), assuming the constant \(c\) is an arbitrary constant of integration. The structure of the particular solution is identical.