Question

Как решить уравнение? y'=(y^2-16)/sin^2(x)

Answer 1

$y'=\frac{y^2-16}{sin^2x}\\\frac{dy}{dx}=\frac{y^2-16}{sin^2x}|*\frac{dx}{y^2-16}\\\frac{dy}{y^2-16}=\frac{dx}{sin^2x}\\\int\frac{dy}{y^2-16}=\int\frac{dx}{sin^2x}\\\\y^2-16=0\\y^2=16\\y=^+_-4\\y'=0\\0=\frac{(^+_-4)^2-16}{sin^2x}\\0=0\\\\\frac{1}{8}ln|\frac{y-4}{y+4}|=-ctg(x)+C\\\frac{1}{8}ln|\frac{y-4}{y+4}|+ctg(x)=C;y=^+_-4$