Question

Знайти частинний розв’язок диференціального рівняння.

Answer 1

Ответ:

1

$y' + \frac{3y}{ x} = \frac{2}{ {x}^{3} } \\ \\ y = uv \\ y' = u'v + v'u \\ \\ u'v + v'u + \frac{3uv}{x} = \frac{2}{ {x}^{3} } \\ u'v + u( v'+ \frac{3v}{x} ) = \frac{2}{ {x}^{3} } \\ \\ 1)v' + \frac{3v}{x} =0 \\ \frac{dv}{dx} = - \frac{3v}{x} \\ \int\limits \frac{dv}{v} = - 3 \int\limits \frac{dx}{x} \\ ln |v| = - 3ln |x| \\ v = \frac{1}{ {x}^{3} } \\ \\ 2)u'v = \frac{2}{ {x}^{3} } \\ \frac{du}{dx} \times \frac{1}{ {x}^{3} } = \frac{2}{ {x}^{3} } \\ u = \int\limits2dx = 2x + C \\ \\ y = \frac{1}{ {x}^{3} } (2x + C) = \frac{2}{ {x}^{2} } + \frac{C}{ {x}^{3} }$

общее решение

$y(1) = 1$

$1 = 2 + C\\ C = - 1$

$y = \frac{2}{ {x}^{2} } - \frac{1}{ {x}^{3} }$

частное решение

2.

$y '- y \cos(x) = \sin(2x) \\ \\ y = uv \\ y' = u'v + v'u \\ \\ u'v + v'u - uv \cos(x) = \sin(2x) \\ u'v + u(v '- v \cos(x)) = \sin(2x) \\ \\ 1) \frac{dv}{dx} = v \cos(x) \\ \int\limits \frac{dv}{v} = \int\limits \cos(x) dx \\ ln(v) = \sin(x) \\ v = {e}^{ \sin(x) } \\ \\ 2)u'v = \sin(2x) \\ \frac{du}{dx} \times {e}^{ \sin(x) } = 2 \sin(x) \cos(x) \\ u = 2 \int\limits {e}^{ - \sin(x) } \sin(x) \cos(x) dx \\ \\ \text{По частям:} \\ U = \sin(x) \: \: \: dU= \cos(x) dx \\ dV = {e}^{ - \sin(x) } \cos(x) dx \: \: \: \\ V= - \int\limits {e}^{ - \sin(x) } d( - \sin(x)) = - {e}^{ - \sin(x) } \\ \\ UV- \int\limits \: VdU = \\ = - \sin( x) e {}^{ \sin(x) } + \int\limits {e}^{ - \sin(x) } \cos(x) dx = \\ = - {e}^{ \sin(x) } \sin(x) - {e}^{ - \sin(x) } + C = \\ = {e}^{ - \sin( x) } ( - \sin(x) - 1) + C \\ \\ y = {e}^{ \sin(x) } \times ( {e}^{ - \sin(x) }( - \sin(x) - 1) + C) \\ y = - \sin(x) - 1 + C {e}^{ \sin(x) }$

общее решение

$y(0) = - 1$

$- 1 = 0 - 1 + C\\ C= 0$

$y = - 1 - \sin(x)$

частное решение

3.

$3(xy'+ y) = x {y}^{2} \\ xy'+ y = \frac{x {y}^{2} }{3} | \div x \\ y'+ \frac{y}{x} = \frac{ {y}^{2} }{3} \\ | \div {y}^{2} \\ \frac{y'}{ {y}^{2} } + \frac{1}{xy} = \frac{1}{3} \\ \\ \frac{1}{y} = z \\ z '= - {y}^{ - 2} \times y' \\ \frac{y'}{ {y}^{2} } = - z '\\ \\ - z' + \frac{z}{x} = \frac{1}{3} \\ z '- \frac{z}{x} = - \frac{1}{3} \\ \\ z = uv \\ z '= u'v + v'u \\ u'v + v'u - \frac{uv}{x} = - \frac{1}{3} \\ u'v + u(v' - \frac{v}{x} ) = - \frac{1}{3} \\ \\ 1) \frac{dv}{dx} = \frac{v}{x} \\ \int\limits \frac{dv}{v} = \int\limits \frac{dx}{x} \\ ln(v) = ln(x) \\ v = x \\ \\ 2) \frac{du}{dx} \times x = - \frac{1}{3} \\ u = - \frac{1}{3} \int\limits \frac{dx}{x} = - \frac{1}{3} ln |x| + C\\ \\ y = x \times ( - \frac{1}{3 ln( |x| ) } + C) \\ y = Cx - \frac{x}{3} ln |x|$

общее решение

$y(1) = 3$

$3 = C- \frac{1}{3} \times ln(1) \\ C= 3$

$y = 3x - \frac{x}{3} ln(x)$

частное решение