Найти общие интегралы уравнений. 1) cos(x)cos(y)dx-sin(x)sin(y)dy=0
2) (x+2)(y^2+1) dx + (y^2-x^2y^2) dy =0
3) (y-1)^2 * dx + (1-x)^3 dy = 0

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Ответ:

odimaa1dimas

16.05.2021 20:56

$\cos(x) \cos(y)dx - \sin(x) \sin(y) dy= 0 \\ \sin(x) \sin(y) dy = \cos(x) \cos(y) dx \\ \int\limits \frac{ \sin(y) }{ \cos(y) } dy = \int\limits\frac{ \sin(x) }{ \cos(x) } dx \\ - \int\limits \frac{d (\cos(y)) }{ \cos(y) } = -\int\limits \frac{d ( \cos(x)) }{ \cos(x) } \\ - ln( | \cos(y) | ) = - ln( | \cos(x) | ) + ln(C) \\ ln( | \cos(y) | ) = ln( | \cos(x) | ) + ln(C) \\ \cos(y) = C\cos(x)$

общее решение

$(x + 2)( {y}^{2} + 1)dx + ( {y}^{2} - {y}^{2} {x}^{2}) dy = 0 \\ {y}^{2} (1 - {x}^{2} )dy = - (x + 2)( {y}^{2} + 1) dx \\ \int\limits \frac{ {y}^{2} }{ {y}^{2} + 1} dy = - \int\limits\frac{x + 2}{1 - {x}^{2} } dx \\ \int\limits \frac{ {y}^{2} + 1 - 1 }{ {y}^{2} + 1} dy = \int\limits \frac{x + 2}{ {x}^{2} - 1} dx \\ \int\limits \: dy - \int\limits \frac{dy}{ {y}^{2} + 1} = \int\limits \frac{xdx}{ {x}^{2} - 1 } + \int\limits \frac{2}{ {x}^{2} - 1 } dx \\ y - arctgy = \frac{1}{2} \int\limits \frac{2xdx}{ {x}^{2} - 1 } + 2 \times \frac{1}{2} ln( | \frac{x - 1}{x + 1} | ) + C \\ y - arctgy = \frac{1}{2} \int\limits \frac{d( {x}^{2} - 1) }{ {x}^{2} - 1 } + ln( | \frac{x - 1}{x + 1} | ) + C \\ y - arctgy = \frac{1}{2} ln( | {x}^{2} - 1 | ) + ln( | \frac{x - 1}{x + 1} | ) + C \\ y - arctgy = ln( | \frac{(x - 1) \sqrt{ {x}^{2} - 1} }{x + 1} | ) + C$

общее решение

$(y - 1) {}^{2} dx + (1 - x) {}^{3} dy = 0 \\ (1 - x) {}^{3} dy = - (y - 1) {}^{2} dx \\ \int\limits \frac{dy}{(y - 1) {}^{2} } = - \int\limits \frac{dx}{(1 - {x})^{3} } \\ \int\limits {(y - 1)}^{ - 2} d(y - 1) = \int\limits {(1 - x)}^{ - 3} d(1 - x) \\ \frac{ {(y - 1)}^{ - 1} }{ - 1} = \frac{ {(1 - x)}^{ - 2} }{ - 2} + C \\ - \frac{1}{y - 1} = - \frac{1}{2 {(1 - x)}^{2} } + C \\ \frac{1}{y - 1} = \frac{1}{2 {(1 - x)}^{2} } + C$