Question #251571

Determine the solution of the following differential equation:

xxy '= 1 + y^2


1
Expert's answer
2021-10-17T15:12:49-0400
xy=1+y2xy '= 1 + y^2

dy1+y2=dxx\dfrac{dy}{1+y^2}=\dfrac{dx}{x}

Integrate


dy1+y2=dxx\int \dfrac{dy}{1+y^2}=\int\dfrac{dx}{x}

tan1y=ln(x)+C\tan^{-1}y=\ln(|x|)+C

y=tan(ln(x)+C)y=\tan(\ln(|x|)+C)


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