How do you find the particular solution to #2xy'-lnx^2=0# that satisfies y(1)=2?

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Answer 1 · 2017-01-25T18:35:16

Use the separation of variables method.
Integrate.
Use #y(1) = 2# to solve for the value of the integration constant.

Separate variables:

#dy = ln(x^2)/(2x)dx#

Integrate:

#y = intln(x^2)/(2x)dx#

#y = intln(x)/(x)dx#

#y = ln^2(x) + C#

Use #y(1) = 2# to solve for c:

#2 = ln^2(1) + C#

#2 = C#

The particular solution is:

#y = ln^2(x) + 2#