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(D) Given the differential equation $\frac{dy}{dx} = (5y+2)(5y-2) = 25y^2 - 4$.Separating the variables,we get $\frac{dy}{25y^2 - 4} = dx$.Using parti

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$0.40$

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(D) Given the differential equation $\frac{dy}{dx} = (5y+2)(5y-2) = 25y^2 - 4$.Separating the variables,we get $\frac{dy}{25y^2 - 4} = dx$.Using partial fractions: $\frac{1}{(5y-2)(5y+2)} = \frac{1}{4} \left( \frac{1}{5y-2} - \frac{1}{5y+2} ight)$.Integrating both sides: $\int \frac{1}{4} \left( \frac{1}{5y-2} - \frac{1}{5y+2} ight) dy = \int dx$.$\frac{1}{4} \left( \frac{1}{5} \ln|5y-2| - \frac{1}{5} \ln|5y+2| ight) = x + C$.$\frac{1}{20} \ln \left| \frac{5y-2}{5y+2} ight| = x + C$.Since $f(0) = 0$,we have $x=0, y=0$: $\frac{1}{20} \ln |\frac{-2}{2}| = 0 + C \Rightarrow C = 0$.So,$\ln \left| \frac{5y-2}{5y+2} ight| = 20x$.$\frac{5y-2}{5y+2} = e^{20x}$.Solving for $y$: $5y-2 = (5y+2)e^{20x} \Rightarrow 5y(1-e^{20x}) = 2(1+e^{20x}) \Rightarrow y = \frac{2}{5} \frac{1+e^{20x}}{1-e^{20x}}$.Wait,checking the sign: $\frac{5y-2}{5y+2} = -e^{20x}$ (since at $x=0, y=0$,$\frac{-2}{2} = -1$).Thus,$\frac{5y-2}{5y+2} = -e^{20x} \Rightarrow 5y-2 = -5ye^{20x} - 2e^{20x} \Rightarrow 5y(1+e^{20x}) = 2(1-e^{20x})$.$y = \frac{2}{5} \frac{1-e^{20x}}{1+e^{20x}}$.As $x ightarrow -\infty$,$e^{20x} ightarrow 0$.Therefore,$\lim_{x ightarrow -\infty} f(x) = \frac{2}{5} 	imes \frac{1-0}{1+0} = \frac{2}{5} = 0.40$.

Let $f: R \rightarrow R$ be a differentiable function with $f(0)=0$. If $y=f(x)$ satisfies the differential equation $\frac{dy}{dx}=(2+5y)(5y-2)$,then the value of $\lim_{x \rightarrow -\infty} f(x)$ is:

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