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

Answer 1

Christopher Allers

#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#

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Answer 2

Aurora Alvarado

The solution to the given differential equation (xy' + 2y = 0) is (y = \frac{c}{x^2}), where (c) is an arbitrary constant.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.