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(A) Let $Y = y+1$. Then $\frac{dY}{dx} = \frac{dy}{dx}$.Substituting into the equation: $\frac{dY}{dx} = Y^{2}e^{x^{2}/2} - xY$.This is a Bernoulli di

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$\frac{-e^{3/2}}{(e^{2}+1)^{2}}$

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(A) Let $Y = y+1$. Then $\frac{dY}{dx} = \frac{dy}{dx}$.Substituting into the equation: $\frac{dY}{dx} = Y^{2}e^{x^{2}/2} - xY$.This is a Bernoulli differential equation. Divide by $Y^{2}$: $Y^{-2}\frac{dY}{dx} + xY^{-1} = e^{x^{2}/2}$.Let $v = Y^{-1} = \frac{1}{y+1}$. Then $\frac{dv}{dx} = -Y^{-2}\frac{dY}{dx}$,so $-\frac{dv}{dx} + xv = e^{x^{2}/2}$,or $\frac{dv}{dx} - xv = -e^{x^{2}/2}$.The integrating factor is $I.F. = e^{\int -x dx} = e^{-x^{2}/2}$.Multiplying by $I.F.$: $\frac{d}{dx}(v e^{-x^{2}/2}) = -1$.Integrating both sides: $v e^{-x^{2}/2} = -x + C$,so $v = (-x+C)e^{x^{2}/2}$.Since $v = \frac{1}{y+1}$,we have $y+1 = \frac{1}{(-x+C)e^{x^{2}/2}}$.Given $y(2)=0$,$1 = \frac{1}{(-2+C)e^{2}}$,so $-2+C = e^{-2}$,which gives $C = 2+e^{-2}$.Thus,$y+1 = \frac{1}{(-x+2+e^{-2})e^{x^{2}/2}}$.At $x=1$,$y+1 = \frac{1}{(-1+2+e^{-2})e^{1/2}} = \frac{1}{(1+e^{-2})e^{1/2}} = \frac{e^{3/2}}{e^{2}+1}$.From the original equation,$y'(1) = (y(1)+1)((y(1)+1)e^{1/2}-1)$.Substituting $y(1)+1 = \frac{e^{3/2}}{e^{2}+1}$: $y'(1) = \frac{e^{3/2}}{e^{2}+1} \left( \frac{e^{3/2}}{e^{2}+1} \cdot e^{1/2} - 1 \right) = \frac{e^{3/2}}{e^{2}+1} \left( \frac{e^{2}}{e^{2}+1} - 1 \right) = \frac{e^{3/2}}{e^{2}+1} \left( \frac{-1}{e^{2}+1} \right) = \frac{-e^{3/2}}{(e^{2}+1)^{2}}$.

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=(y+1)((y+1)e^{x^{2}/2}-x)$,with $y(2)=0$. Then $y'(1)$ is equal to . . . .

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