Question

Помогите решить Дифференциальные уравнения

Answer 1

Ответ:

1

$xdy = ydx \\ \int\limits \frac{dy}{y} = \int\limits \frac{dx}{x} \\ ln |y| = ln |x| + ln |C| \\ y = Cx$

общее решение

2

$4y' = \frac{ {y}^{2} }{ {x}^{2} } + 10 \frac{y}{x} + 5 \\ \\ \frac{y}{x} = u \\ y' = u'x + u \\ \\ 4(u'x + u) = {u}^{2} + 10 u + 5 \\ u'x + u = \frac{ {u}^{2} + 10u + 5}{4} \\ \frac{du}{dx} x = \frac{ {u}^{2} + 10 u + 5 - 4u}{4} \\ \frac{du}{dx} x = \frac{ {u}^{2} + 6u + 5}{4} \\ \int\limits \frac{4dy}{u {}^{2} + 6 u + 5} = \int\limits \frac{dx}{x} \\ \\ {u}^{2} + 6u + 5 = {u}^{2} + 2 \times u \times 3 + 9 - 4 = \\ = {(u + 3)}^{2} - 4 = {(u + 3)}^{2} - {2}^{2} \\ \\ \int\limits \frac{4du}{ {(u + 1)}^{2} - {2}^{2} } = ln |x| + C \\ 4\int\limits \frac{d(u +1 )}{ {(u + 1)}^{2} - {2}^{2} } = ln |x| + C \\ \frac{4}{2 \times 2} ln | \frac{u + 1 - 2}{u + 1 + 2} | = ln |x| + C\\ ln | \frac{u - 1}{u + 3} | = ln |x| + ln |C| \\ \frac{u - 1}{u + 3} = Cx \\ \frac{ \frac{y}{x} - 1}{ \frac{y}{x} + 3} = Cx \\ \frac{y - x}{x} \times \frac{x}{y + 3x} = Cx \\ \frac{y - x}{3x + 4} = Cx$

общее решение

3

$y'' - 2y' + 2y = 0 \\ \\ y = {e}^{kx} \\ {e}^{kx}( k {}^{2} - 2 k + 2) = 0 \\ D = 4 - 8 = - 4 \\ k_1 = \frac{2 + \sqrt{ - 4} }{2} = \frac{2 + 2i}{2} = 1 + i \\ k_2 = 1 - i \\ y =e {}^{x} ( C_1 \sin(x) + C_2 \cos(x))$

общее решение

$y(0) = 1,y'(0) = 3$

$y' = e {}^{x} (C_1 \sin(x) + C_2 \cos(x)) + {e}^{x} (C_1\cos(x) - C_2 \sin(x))$

$\left \{ {{C_1 \sin(0) + C_2 \cos(0) = 1 } \atop {C_1 \sin(0) + C_2 \cos(0) + C_1 \cos(0) - C_2 \sin(0) = 3 } } \right. \\ \\ \left \{ {{C_2 = 1} \atop {C_2 + C_1 = 3} } \right. \\ \\ \left \{ {{C_2 = 1} \atop {C_1 = 2} } \right. \\ \\ y = {e}^{x} ( 2\sin(x) + \cos(x))$

частное решение