text: solve: y" - (1/x)*y' = 0? this is for my differential equations class. y = cx^2/2 + c1   {c and c1 constants}  let z = y'.  then z' = y'' rewrite equation as: z' - (1/x)z = 0 z' = z/x z'/z = 1/x  ∫ z'/z dx = ∫ 1/x dx ∫ (1/z)(dz/dx)dx = ∫ 1/x dx ∫ 1/z dz = ∫ 1/x dx ln(z) = ln(x) + c0 {c0 constant} z = e^(c0)*x = cx  y' = cx y = cx^2/2 + c1