Ответ:
y=-1/(sinx+C)
Объяснение:
[tex]y'=y^2cosx\\\\\\\frac{dy}{dx} =y^2cos(x)\\\\\frac{dy}{y^2}=cos(x)dx \\\\\int\limits^{}_{} {\frac{1}{y^2} } \, dy = \int\limits^{}_{}cos(x)dx\\-\frac{1}{y} = sin(x)+C\\\\y= -\frac{1}{sin(x)+C}[/tex]
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Ответ:
y=-1/(sinx+C)
Объяснение:
[tex]y'=y^2cosx\\\\\\\frac{dy}{dx} =y^2cos(x)\\\\\frac{dy}{y^2}=cos(x)dx \\\\\int\limits^{}_{} {\frac{1}{y^2} } \, dy = \int\limits^{}_{}cos(x)dx\\-\frac{1}{y} = sin(x)+C\\\\y= -\frac{1}{sin(x)+C}[/tex]