Let the solution curve y=y(x) of the differen | vedclass

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(A) The given differential equation is $(4+x^{2}) dy = 2x(x^{2}+3y+4) dx$.Rearranging,we get $(x^{2}+4) \frac{dy}{dx} = 2x^{3} + 6xy + 8x$.$(x^{2}+4)

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$8$

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(A) The given differential equation is $(4+x^{2}) dy = 2x(x^{2}+3y+4) dx$.Rearranging,we get $(x^{2}+4) \frac{dy}{dx} = 2x^{3} + 6xy + 8x$.$(x^{2}+4) \frac{dy}{dx} - 6xy = 2x(x^{2}+4)$.Dividing by $(x^{2}+4)$,we get $\frac{dy}{dx} - \frac{6x}{x^{2}+4} y = 2x$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{6x}{x^{2}+4}$ and $Q(x) = 2x$.The integrating factor $I.F. = e^{\int P(x) dx} = e^{-\int \frac{6x}{x^{2}+4} dx} = e^{-3 \ln(x^{2}+4)} = (x^{2}+4)^{-3} = \frac{1}{(x^{2}+4)^{3}}$.The solution is $y \cdot I.F. = \int Q(x) \cdot I.F. dx + C$.$y \cdot \frac{1}{(x^{2}+4)^{3}} = \int \frac{2x}{(x^{2}+4)^{3}} dx$.Let $t = x^{2}+4$,then $dt = 2x dx$.$y \cdot \frac{1}{(x^{2}+4)^{3}} = \int t^{-3} dt = \frac{t^{-2}}{-2} + C = -\frac{1}{2(x^{2}+4)^{2}} + C$.Since the curve passes through $(0,0)$,we have $0 = -\frac{1}{2(4)^{2}} + C \implies C = \frac{1}{32}$.Thus,$y = -\frac{(x^{2}+4)^{3}}{2(x^{2}+4)^{2}} + \frac{(x^{2}+4)^{3}}{32} = -\frac{x^{2}+4}{2} + \frac{(x^{2}+4)^{3}}{32}$.For $x=2$,$y(2) = -\frac{4+4}{2} + \frac{(4+4)^{3}}{32} = -4 + \frac{512}{32} = -4 + 16 = 8$.

Let the solution curve $y=y(x)$ of the differential equation $(4+x^{2}) dy - 2x(x^{2}+3y+4) dx = 0$ pass through the origin. Then $y(2)$ is equal to

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