If the solution of the differential equation | vedclass

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(D) The given differential equation is $(1 + x^3) \frac{dy}{dx} + 6x^2y = 1 + x^2$. Dividing by $(1 + x^3)$, we get $\frac{dy}{dx} + \frac{6x^2}{1 + x

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) The given differential equation is $(1 + x^3) \frac{dy}{dx} + 6x^2y = 1 + x^2$. Dividing by $(1 + x^3)$, we get $\frac{dy}{dx} + \frac{6x^2}{1 + x^3} y = \frac{1 + x^2}{1 + x^3}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{6x^2}{1 + x^3}$ and $Q(x) = \frac{1 + x^2}{1 + x^3}$.The integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{6x^2}{1 + x^3} dx} = e^{2 \ln(1 + x^3)} = (1 + x^3)^2$.The solution is $y \cdot IF = \int Q(x) \cdot IF dx + c$.$y(1 + x^3)^2 = \int \frac{1 + x^2}{1 + x^3} (1 + x^3)^2 dx + c = \int (1 + x^2)(1 + x^3) dx + c = \int (1 + x^3 + x^2 + x^5) dx + c$.$y(1 + x^3)^2 = x + \frac{x^4}{4} + \frac{x^3}{3} + \frac{x^6}{6} + c$.Comparing this with $y = \frac{1}{(1 + x^3)^s} [x + \frac{x^p}{p} + \frac{x^q}{q} + \frac{x^r}{r} + c]$, we get $s = 2$ and the set $\{p, q, r\} = \{4, 3, 6\}$.The $LCM$ of $2, 3, 4, 6$ is $12$.

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...

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