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(D) Given the differential equation $\frac{d y}{d x}=2 x(x+y)^3-x(x+y)-1$. Let $t = x+y$,then $\frac{d t}{d x} = 1 + \frac{d y}{d x}$,so $\frac{d y}{d

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$\frac{1}{2-\sqrt{e}}$

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(D) Given the differential equation $\frac{d y}{d x}=2 x(x+y)^3-x(x+y)-1$. Let $t = x+y$,then $\frac{d t}{d x} = 1 + \frac{d y}{d x}$,so $\frac{d y}{d x} = \frac{d t}{d x} - 1$. Substituting this into the equation: $\frac{d t}{d x} - 1 = 2xt^3 - xt - 1$,which simplifies to $\frac{d t}{d x} = 2xt^3 - xt = x(2t^3 - t)$. Separating variables: $\frac{d t}{2t^3 - t} = x dx$. Using partial fractions: $\frac{1}{t(2t^2 - 1)} = \frac{A}{t} + \frac{Bt+C}{2t^2 - 1}$. Solving gives $A = -1, B = 2, C = 0$. So,$\int (\frac{2t}{2t^2 - 1} - \frac{1}{t}) dt = \int x dx$. Integrating: $\frac{1}{2} \ln|2t^2 - 1| - \ln|t| = \frac{x^2}{2} + C$. At $x=0, y=1$,so $t = 0+1 = 1$. $\frac{1}{2} \ln|2(1)^2 - 1| - \ln|1| = 0 + C \implies C = 0$. Thus,$\ln|\frac{\sqrt{2t^2 - 1}}{t}| = \frac{x^2}{2}$. Squaring both sides: $\frac{2t^2 - 1}{t^2} = e^{x^2}$. For $x = \frac{1}{\sqrt{2}}$,$x^2 = \frac{1}{2}$,so $\frac{2t^2 - 1}{t^2} = e^{1/2} = \sqrt{e}$. $2t^2 - 1 = t^2 \sqrt{e} \implies t^2(2 - \sqrt{e}) = 1 \implies t^2 = \frac{1}{2 - \sqrt{e}}$. Since $t = x+y$,the value is $\frac{1}{2 - \sqrt{e}}$.

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}=2 x(x+y)^3-x(x+y)-1$,with the initial condition $y(0)=1$. Then,the value of $\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2$ is equal to:

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