Question

If the solution of the differential equation \((1+x^3)\frac{dy}{dx}+6x^2y = 1+x^2\) is \(y = \frac{1}{(1+x^3)^s}[x+\frac{x^p}{p}+\frac{x^q}{q}+\frac{x^r}{r}+c]\), then the LCM of \(p,q,r\) and \(s\) is...

• A. \(1\)
• B. \(6\)
• C. \(4\)
• D. \(12\)

Hint:

The integrating factor is (1 + x^3)^2, since the coefficient of y over the coefficient of y-prime is 6x^2/(1 + x^3).

Answer 1

The Correct Option is D

Step 1: Convert to standard form
Divide by \((1 + x^3)\): \(\dfrac{dy}{dx} + \dfrac{6x^2}{1 + x^3}y = \dfrac{1 + x^2}{1 + x^3}\).

Step 2: Integrating factor
\[ \text{IF} = e^{\int\frac{6x^2}{1+x^3}dx} = e^{2\ln(1 + x^3)} = (1 + x^3)^2 \]

Step 3: Solve
\[ y(1 + x^3)^2 = \int\frac{1 + x^2}{1 + x^3}(1 + x^3)^2\,dx = \int(1 + x^2)(1 + x^3)\,dx = \int(1 + x^2 + x^3 + x^5)\,dx \]
This is \(x + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^6}{6} + c\).

Step 4: Compare
Hence \(s = 2\) and \(\{p, q, r\} = \{3, 4, 6\}\). The LCM of \(3, 4, 6, 2\) is 12, option (D).

Final Answer:
The LCM is 12. This is option (D). \[ \boxed{\text{(D) }12} \]