If the solution of the differential equation | vedclass

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(D) Step $1$: Rewrite the equation in standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.$\frac{dy}{dx} + \frac{6x^2}{1+x^3}y = \frac{1+x^2}{1+x^3}$.

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) Step $1$: Rewrite the equation in standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.$\frac{dy}{dx} + \frac{6x^2}{1+x^3}y = \frac{1+x^2}{1+x^3}$.Step $2$: Find the integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{6x^2}{1+x^3} dx} = e^{2 \ln(1+x^3)} = (1+x^3)^2$.Step $3$: Multiply by $IF$ and integrate: $y(1+x^3)^2 = \int (1+x^2)(1+x^3) dx = \int (1 + x^2 + x^3 + x^5) dx = x + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^6}{6} + c$.Step $4$: Comparing with $y = \frac{1}{(1+x^3)^2} [x + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^6}{6} + c]$, we get $s=2, p=3, q=4, r=6$.Step $5$: 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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