Question

Решение дифференциальных уравнений:
1. (x^2+y^2)dx-xydy=0
2. y'=tg3y*Inx
3. y'(1-x^2)^(1/2)+y=arcsin x

Answer 1

Ответ:

1

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

общее решение

2

$y '= tg(3y) \times ln(x) \\ \frac{dy}{dx} = tg(3y) \times ln(x) \\ \int\limits \frac{dy}{tg(3y)} = \int\limits ln(x) dx \\ \\ 1)\int\limits \frac{dy}{tg(3y)} = \frac{1}{3} \int\limits \frac{d(3y)}{tg(3y)} = \\ = \frac{1}{3} \int\limits \frac{ \cos(3y) }{ \sin(3y) } d(3y) = \frac{1}{3} \int\limits \frac{d( \sin(3y)) }{ \sin(3y) } = \\ = \frac{1}{3} ln | \sin(3y) | + C \\ \\ 2)\int\limits ln(x) dx \\ \\ \text{По частям:} \\ U= ln(x) \: \: \: \: dU = dx \\ dV = \frac{dx}{x} \: \: \: \: \: \: V = x \\ \\ UV- \int\limits \: VdU = \\ = x ln(x) - \int\limits \: xdx \times \frac{1}{x} = \\ = x ln(x) - x + C = x( ln(x) - 1) + C \\ \\ \frac{1}{3} ln | \sin(3y) | = x( ln(x) - 1) + C\\ ln | \sin(3y) | = 3x (ln(x) - 1) + C$

общее решение

3.

$y' \sqrt{1 - {x}^{2} } + y = arcsinx \: \: \: | \div \sqrt{1 - {x}^{2} } \\ y' + \frac{y}{ \sqrt{1 - {x}^{2} } } = \frac{arcsinx}{ \sqrt{1 - {x}^{2} } } \\ \\ y = uv \\ y '= u'v + v'u \\ \\ u'v + v'u + \frac{uv}{ \sqrt{1 - {x}^{2} } } = \frac{arcsinx}{ \sqrt{1 - {x}^{2} } } \\ u'v + u(v' + \frac{v}{ \sqrt{1 - {x}^{2} } } ) = \frac{arcsinx}{ \sqrt{1 - {x}^{2} } } \\ \\ 1)v' + \frac{v}{ \sqrt{1 - {x}^{2} } } \\ \frac{dv}{dx} = - \frac{v}{ \sqrt{1 - {x}^{2} } } \\ \int\limits \frac{dv}{v} = - \int\limits \frac{dx}{ \sqrt{1 - {x}^{2} } } \\ ln |v| = - arcsinx \\ v = {e}^{ - arcsinx} \\ \\ 2)u'v = \frac{arcsinx}{ \sqrt{1 - {x}^{2} } } \\ \frac{du}{dx} \times {e}^{ - arcsinx} = \frac{arcsinx}{ \sqrt{1 - {x}^{2} } } \\ u = \int\limits {e}^{arcsinx} \frac{arcsinx}{ \sqrt{1 - {x}^{2} } } dx \\ \\ \text{По частям:} \\ U= arcsinx \: \: \: \: \: \: \: dU = \frac{dx}{ \sqrt{1 - {x}^{2} } } \\ dV = {e}^{arcsinx} \frac{dx}{ \sqrt{1 - {x}^{2} } } \: \: \: V = \int\limits {e}^{arcsinx} d(arcsinx) = {e}^{arcsinx} \\ \\ {e}^{arcsinx} arcsinx - \int\limits {e}^{arcsinx} \times \frac{dx}{ \sqrt{1 - {x}^{2} } } = \\ = {e}^{arcsinx}arcsinx - {e}^{arsinx} + C = \\ = {e}^{arcsinx} (arcsinx - 1) + C\\ \\ \\ \\2) u = {e}^{arcinx}( arcsinx - 1) + C \\ \\ y = uv = \\ = {e}^{ - arcsinx} \times ( {e}^{arcsinx}( arcsinx - 1) + C) \\ y = arcsinx - 1 + C {e}^{ - arcsinx}$

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