A curve passes through the point (x = 1, y = 0) and satisfies the differential equation $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{{\rm{x}}^2} + {{\rm{y}}^2}}}{{2{\rm{y}}}} + \frac{{\rm{y}}}{{\rm{x}}}.$ The equation that describes the curve is

Question

A curve passes through the point (x = 1, y = 0) and satisfies the differential equation $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{{\rm{x}}^2} + {{\rm{y}}^2}}}{{2{\rm{y}}}} + \frac{{\rm{y}}}{{\rm{x}}}.$ The equation that describes the curve is

TuteeHUB · Accepted Answer

$\frac{1}{2}{\rm{In}}\left( {1 + \frac{{{{\rm{y}}^2}}}{{{{\rm{x}}^2}}}} \right) = {\rm{x}} - 1$

TuteeHUB · Answer

${\rm{In}}\left( {1 + \frac{{{{\rm{y}}^2}}}{{{{\rm{x}}^2}}}} \right) = {\rm{x}} - 1$

TuteeHUB · Answer

${\rm{In}}\left( {1 + \frac{{\rm{y}}}{{\rm{x}}}} \right) = {\rm{x}} - 1$

TuteeHUB · Answer

$\frac{1}{2}{\rm{In}}\left( {1 + \frac{{\rm{y}}}{{\rm{x}}}} \right) = {\rm{x}} - 1$

1.	A curve passes through the point (x = 1, y = 0) and satisfies the differential equation \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{{\rm{x}}^2} + {{\rm{y}}^2}}}{{2{\rm{y}}}} + \frac{{\rm{y}}}{{\rm{x}}}.\) The equation that describes the curve is
A.	\({\rm{In}}\left( {1 + \frac{{{{\rm{y}}^2}}}{{{{\rm{x}}^2}}}} \right) = {\rm{x}} - 1\)
B.	\(\frac{1}{2}{\rm{In}}\left( {1 + \frac{{{{\rm{y}}^2}}}{{{{\rm{x}}^2}}}} \right) = {\rm{x}} - 1\)
C.	\({\rm{In}}\left( {1 + \frac{{\rm{y}}}{{\rm{x}}}} \right) = {\rm{x}} - 1\)
D.	\(\frac{1}{2}{\rm{In}}\left( {1 + \frac{{\rm{y}}}{{\rm{x}}}} \right) = {\rm{x}} - 1\)
Answer» B. \(\frac{1}{2}{\rm{In}}\left( {1 + \frac{{{{\rm{y}}^2}}}{{{{\rm{x}}^2}}}} \right) = {\rm{x}} - 1\)

A curve passes through the point (x = 1, y = 0) and satisfies the differential equation \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{{\rm{x}}^2} + {{\rm{y}}^2}}}{{2{\rm{y}}}} + \frac{{\rm{y}}}{{\rm{x}}}.\) The equation that describes the curve is

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