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Quiz-1 / TUT0402 Quiz1
« on: September 27, 2019, 02:12:57 PM »
xy' = (1-y^2)^(1/2)
solution: x(dy/dx) = (1-y^2)^(1/2)
∫ 1/ (1-y^2)^(1/2) dy = ∫1/x dx
arcsin(y) = ln|x| + C
y = sin(ln|x| + C)
solution: x(dy/dx) = (1-y^2)^(1/2)
∫ 1/ (1-y^2)^(1/2) dy = ∫1/x dx
arcsin(y) = ln|x| + C
y = sin(ln|x| + C)