What is a solution to the differential equation #xy' + 2y = 0#?

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Answer 1 · 2016-09-08T04:00:42

#y=C/x^2#

Write #y'# as #dy/dx#:

#xdy/dx+2y=0#

Try to separate the variables:

#xdy/dx=-2y#

#-1/(2y)dy=1/xdx#

Integrate both sides:

#-1/2int1/ydy=int1/xdx#

#-1/2lny=lnx+C#

Remember that any modification to #C# is basically immaterial, it stays #C# because it becomes some other unimportant constant:

#lny=-2lnx+C#

#y=e^(-2lnx+C#

#y=e^(-2lnx)*e^C#

#y=Ce^(ln(x^-2))#

#y=Cx^-2#

#y=C/x^2#