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(B) Given differential equation is $(1+e^{x}) dy+(1+y^{2}) e^{x} dx=0$.Separating the variables,we get $\frac{dy}{1+y^{2}} + \frac{e^{x} dx}{1+e^{x}}

Vedclass · Accepted Answer

$	an ^{-1} y+	an ^{-1}(e^{x})=\frac{\pi}{4}$

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(B) Given differential equation is $(1+e^{x}) dy+(1+y^{2}) e^{x} dx=0$.Separating the variables,we get $\frac{dy}{1+y^{2}} + \frac{e^{x} dx}{1+e^{x}} = 0$.Integrating both sides,we get $\int \frac{dy}{1+y^{2}} + \int \frac{e^{x} dx}{1+e^{x}} = C$.Let $1+e^{x} = t$,then $e^{x} dx = dt$.So,$	an^{-1} y + \int \frac{dt}{t} = C$.$	an^{-1} y + \ln|1+e^{x}| = C$.Given $y=1$ at $x=0$,we have $	an^{-1}(1) + \ln|1+e^{0}| = C$.$\frac{\pi}{4} + \ln(2) = C$.Thus,the particular solution is $	an^{-1} y + \ln(1+e^{x}) = \frac{\pi}{4} + \ln(2)$.

Find the particular solution of the differential equation $(1+e^{x}) dy+(1+y^{2}) e^{x} dx=0$,given that $y=1$ when $x=0$.

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