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: Spot the exact derivative
$\dfrac{d}{dx}\left[(1 + x^3)^2y\right] = (1 + x^3)\left[(1 + x^3)y' + 6x^2y\right]$.

Step 2: Use the equation
The bracket equals $1 + x^2$, so $\dfrac{d}{dx}\left[(1 + x^3)^2y\right] = (1 + x^2)(1 + x^3)$.

Step 3: Integrate
$(1 + x^3)^2y = x + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^6}{6} + c$, so $s = 2$ and the denominators are 3, 4 and 6.

Step 4: LCM
$\text{LCM}(3, 4, 6, 2) = 12$. The values 1, 4 and 6 would not be divisible by all of these numbers.

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