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(C) The given differential equation is $(1-x^2) dy = [xy + (x^3+2) \sqrt{3(1-x^2)}] dx$.Dividing by $(1-x^2) dx$,we get $\frac{dy}{dx} - \frac{x}{1-x^

Vedclass · Accepted Answer

$97$

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(C) The given differential equation is $(1-x^2) dy = [xy + (x^3+2) \sqrt{3(1-x^2)}] dx$.Dividing by $(1-x^2) dx$,we get $\frac{dy}{dx} - \frac{x}{1-x^2} y = \frac{(x^3+2) \sqrt{3(1-x^2)}}{1-x^2}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{x}{1-x^2}$ and $Q(x) = \frac{(x^3+2) \sqrt{3}}{\sqrt{1-x^2}}$.The integrating factor $IF = e^{\int P(x) dx} = e^{-\int \frac{x}{1-x^2} dx} = e^{\frac{1}{2} \ln(1-x^2)} = \sqrt{1-x^2}$.The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.$y \sqrt{1-x^2} = \int \frac{(x^3+2) \sqrt{3}}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} dx + C = \sqrt{3} \int (x^3+2) dx + C$.$y \sqrt{1-x^2} = \sqrt{3} (\frac{x^4}{4} + 2x) + C$.Given $y(0)=0$,we have $0 = \sqrt{3}(0) + C$,so $C=0$.Thus,$y \sqrt{1-x^2} = \sqrt{3} (\frac{x^4}{4} + 2x)$.For $x = 1/2$,$y \sqrt{1 - 1/4} = \sqrt{3} (\frac{(1/2)^4}{4} + 2(1/2)) = \sqrt{3} (\frac{1}{64} + 1) = \sqrt{3} (\frac{65}{64})$.$y \sqrt{3/4} = y \frac{\sqrt{3}}{2} = \sqrt{3} \frac{65}{64}$.$y = \frac{65}{64} 	imes 2 = \frac{65}{32}$.Since $m=65$ and $n=32$ are coprime,$m+n = 65+32 = 97$.

Let $y=y(x)$ be the solution of the differential equation $(1-x^2) dy = [xy + (x^3+2) \sqrt{3(1-x^2)}] dx$ for $-1 < x < 1$,with $y(0)=0$. If $y(1/2) = m/n$,where $m$ and $n$ are coprime numbers,then $m+n$ is equal to . . . . . . . . . .

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