Solve the initial value problem y'=ky^2\lnx, with y(1)=-1?
My work:
dy/dx=ky^2\lnx
dy/y^2=k\lnxdx
\inty^-2dy=\intk\lnxdx=k\int\lnxdx
See answer below for the rest
My work:
See answer below for the rest
1 Answer
May 30, 2018
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Explanation:
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