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(A) The given differential equation is $(x^2-4)y^{\prime}-2xy = -2x(4-x^2)^2$. Dividing by $(x^2-4)^2$,we get $\frac{(x^2-4)y^{\prime}-2xy}{(x^2-4)^2}

Vedclass · Accepted Answer

$16$

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(A) The given differential equation is $(x^2-4)y^{\prime}-2xy = -2x(4-x^2)^2$. Dividing by $(x^2-4)^2$,we get $\frac{(x^2-4)y^{\prime}-2xy}{(x^2-4)^2} = -2x$. This is the derivative of the quotient: $\frac{d}{dx} \left( \frac{y}{x^2-4} ight) = -2x$. Integrating both sides with respect to $x$: $\frac{y}{x^2-4} = -x^2 + C$. So,$y = (-x^2+C)(x^2-4)$. Given that the curve passes through $(3, 15)$,we substitute $x=3$ and $y=15$: $15 = (-9+C)(9-4) \Rightarrow 15 = 5(-9+C) \Rightarrow 3 = -9+C \Rightarrow C=12$. Thus,$f(x) = (12-x^2)(x^2-4)$. To find the local maximum,we set $f^{\prime}(x) = 0$: $f^{\prime}(x) = (-2x)(x^2-4) + (12-x^2)(2x) = -2x^3 + 8x + 24x - 2x^3 = -4x^3 + 32x = 0$. $-4x(x^2-8) = 0$. Since $x>2$,we have $x^2=8$,so $x=2\sqrt{2}$. The local maximum value is $f(2\sqrt{2}) = (12-8)(8-4) = 4 	imes 4 = 16$.

If the solution curve $y=f(x)$ of the differential equation $(x^{2}-4)y^{\prime}-2xy+2x(4-x^{2})^{2}=0$ for $x>2$ passes through the point $(3, 15)$,then the local maximum value of $f$ is:

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