Question

How do I find two linear solutions for the linear equation: y" + (1/x)y' = 0 ?

Answer 1

Hi hollyn,

If you make a substitution\[v=y'\]then\[v'=y''\]and your equation becomes\[v'+\frac{1}{x}v=0\]This equation is separable,\[\frac{dv}{dx}=-\frac{v}{x} \rightarrow \frac{dv}{v}=-\frac{dx}{x}\]Integrate both sides\[\ln v = - \ln x +c \]where c is some constant.Exponentiating both sides leaves you with\[v=\frac{c_1}{x}\]where c_1 is some constant. But remember, \[v=y'\]so\[\frac{dy}{dx}=\frac{c_1}{x} \rightarrow y=c_1 \ln x + c_2\]where c_2 is an arbitrary constant.