How to you find the general solution of #dy/dx=x/y#?

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How to you find the general solution of #dy/dx=x/y#?

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Answer 1 · 2017-01-13T08:47:46

#x^2-y^2=c#

Explanation:

#dy/{dx}=x/y#

#ydy=xdx# by exploiting the notation (separation)

#int ydy=int xdx# further exploiting the notation

#1/2y^2=1/2x^2+d#
#y^2=x^2+2d#
#x^2-y^2=-2d#
#x^2-y^2=c# where #c=-2d#
Depending on whether #c# is positive, negative or zero you get a hyperbola open to the #x#-axis, open to the #y#=axis, or a pair of straight lines through the origin.
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How to you find the general solution of #dy/dx=x/y#?

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