Question

Помогите пожалуйста проинтегрировать уравнения

Answer 1

Ответ:

1.

$y'ctgx + y = 2 \: \: \: | \div ctgx \\ y' + \frac{y}{ctgx} = \frac{2}{ctgx} \\ \\ y = uv \\ y' = u'v + v'u \\ \\ u'v + v'u + \frac{uv}{ctgx} = \frac{2}{ctgx} \\ u' v + u(v' + \frac{v}{ctgx} ) = \frac{2}{ctgx} \\ \\ 1)v' + \frac{v}{ctgx} = 0 \\ \frac{dv}{dx} = -v tgx \\ \int\limits \frac{dv}{v} = - \int\limits \frac{ \sin(x) }{ \cos(x) } dx \\ ln(v) = \int\limits \frac{d (\cos(x)) }{ \cos(x) } \\ ln(v) = ln( \cos(x) ) \\ v = \cos(x) \\ \\ 2)u'v = 2tgx \\ \frac{du}{dx} \times \cos(x) = 2 \times \frac{ \sin(x) }{ \cos(x) } \\ \int\limits \: du = 2\int\limits \frac{ \sin( x) }{ \cos {}^{2} ( x) } dx \\ u = - 2\int\limits \frac{d( \cos(x)) }{ \cos {}^{2} (x) } = \\ = - 2 \times \frac{ { (\cos(x)) }^{ - 1} }{ - 1} + C = \frac{2}{ \cos(x) } + C \\ \\ y = \cos(x) \times ( \frac{2}{ \cos(x) } + C) \\ y = 2 + \frac{C}{ \cos(x) }$

общее решение

2.

$\frac{1}{ \cos {}^{2} (x) } \times ctgydx + \frac{1}{ \cos {}^{2} (y) } tgxdy = 0 \\ \frac{tgx}{ \cos {}^{2} (y) }dy = - \frac{ctgy}{ \cos {}^{2} (x) } dx \\ \int\limits \frac{dy}{ctgy \times \cos {}^{2} (y) } = - \int\limits \frac{dx}{tgx \times \cos {}^{2} (x) } \\ \int\limits \frac{1}{ \cos {}^{2} (y) } \times tgydy = - \int\limits \frac{1}{ \cos {}^{2} (x) } \times \frac{1}{tgx} dx \\ \int\limits \: tgyd(tgy) = - \int\limits \frac{d(tgx)}{tgx} \\ \frac{{tg}^{2} y}{2} = ln |tgx| + C\\ {tg}^{2} y = 2 ln |tgx| + C$

общее решение

3.

$xy' = y( ln(y) - ln(x)) \: \: \: | \div x \\ y'= \frac{y}{x} ln( \frac{y}{x} ) \\ \\ \frac{y}{x} = u \\ y' = u'x + u \\ \\ u'x + u = u ln(u) \\ \frac{du}{dx} x = u ln(u) - u \\ \int\limits \frac{du}{u (ln(u) - 1) } = \int\limits \frac{dx}{x} \\ \int\limits \frac{d (ln(u) )}{ ln(u) - 1 } = ln(x) + ln(C) \\ \int\limits \frac{d (ln(u) - 1) }{ ln(u) - 1} = ln(Cx) \\ ln( ln(u) - 1) = ln(Cx) \\ ln(u) - 1 = Cx \\ ln( \frac{y}{x} ) = Cx + 1$

общее решение

4.

$( {x}^{2} + 2 {y}^{2} )dx + 2xydy = 0 \\ 2xydy = - ( {x}^{2} + 2 {y}^{2} )dx \\ 2xyy' = - ( {x}^{2} + 2 {y}^{2} ) \: \: \: | \div {x}^{2} \\ 2 \frac{y}{x} \times y' = - 1 - 2 \frac{ {y}^{2} }{ {x}^{2} } \\ \\ \frac{y}{x} = u \\ y' = u'x + u \\ \\ 2u \times (u'x + u) = - 1 - 2 {u}^{2} \\ u'x + u = \frac{ - 1 - 2 {u}^{2} }{2u} \\ \frac{du}{dx} x = \frac{ - 1 - 2 {u}^{2} - 2 {u}^{2} }{2u} \\ \frac{du}{dx} x = \frac{ - 4 {u}^{2} - 1}{2u} \\ - \int\limits \frac{2udu}{4u {}^{2} + 1 } = \int\limits \frac{dx}{x} \\ - \frac{1}{4} \int\limits \frac{8udu}{4u {}^{2} + 1} = ln |x| + ln |C| \\ - \frac{1}{4} \int\limits \frac{d(4u {}^{2} + 1)}{4u {}^{2} + 1 } = ln(Cx) \\ - \frac{1}{4} ln | 4{u}^{2} + 1 | = ln(Cx) \\ ln |4 {u}^{2} + 1 | = - 4 ln(Cx) \\ 4 {u}^{2} + 1 = \frac{1}{C {x}^{4} } \\ \frac{4 {y}^{2} }{ {x}^{2} } + 1 = \frac{1}{Cx {}^{4} }$

общее решение