The solution of the differential equation (1+ | vedclass

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(D) Given the differential equation: $(1+e^{-x})(1+y^2) \frac{dy}{dx} = y^2$ Separate the variables: $\frac{1+y^2}{y^2} dy = \frac{1}{1+e^{-x}} dx$ Si

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$y^2=1+y \log (\frac{1+e^x}{2})$

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(D) Given the differential equation: $(1+e^{-x})(1+y^2) \frac{dy}{dx} = y^2$ Separate the variables: $\frac{1+y^2}{y^2} dy = \frac{1}{1+e^{-x}} dx$ Since $\frac{1}{1+e^{-x}} = \frac{e^x}{e^x+1}$,we have: $\int (y^{-2} + 1) dy = \int \frac{e^x}{e^x+1} dx$ Integrating both sides: $-\frac{1}{y} + y = \log(1+e^x) + C$ Multiply by $y$: $y^2 - 1 = y \log(1+e^x) + Cy$ $y^2 - 1 = y(\log(1+e^x) + C)$ Given the curve passes through $(0,1)$,substitute $x=0, y=1$: $1^2 - 1 = 1(\log(1+e^0) + C) \Rightarrow 0 = \log(2) + C \Rightarrow C = -\log(2)$ Substitute $C$ back into the equation: $y^2 - 1 = y(\log(1+e^x) - \log(2))$ $y^2 - 1 = y \log(\frac{1+e^x}{2})$ $y^2 = 1 + y \log(\frac{1+e^x}{2})$

The solution of the differential equation $(1+e^{-x})(1+y^2) \frac{dy}{dx} = y^2$ which passes through the point $(0,1)$ is

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