The solution curve of the differential equati | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given 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$$\Rightarrow (y^{-2}+1) dy = \frac{e^{x}}{e^{x}+1} dx$.Integrating both sides:$\int (y^{-2}+1) dy = \int \frac{e^{x}}{e^{x}+1} dx$$-y^{-1} + y = \ln(e^{x}+1) + C$$y - \frac{1}{y} = \ln(e^{x}+1) + C$.Since the curve passes through $(0,1)$,substitute $x=0$ and $y=1$:$1 - \frac{1}{1} = \ln(e^{0}+1) + C$$0 = \ln(2) + C \Rightarrow C = -\ln(2)$.Substituting $C$ back into the equation:$y - \frac{1}{y} = \ln(e^{x}+1) - \ln(2)$$y - \frac{1}{y} = \ln\left(\frac{e^{x}+1}{2}\right)$Multiply by $y$:$y^{2} - 1 = y \ln\left(\frac{1+e^{x}}{2}\right)$$y^{2} = 1 + y \ln\left(\frac{1+e^{x}}{2}\right)$.

The solution curve 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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