Question

Решить Дифференциальные уравнения

Answer 1

а)

$2xy' + {y}^{2} = 1 \\ 2x \frac{dy}{dx} = 1 - {y}^{2} \\ \int\limits \frac{dy}{1 - {y}^{2} } = \int\limits \frac{dx}{2x} \\ arcsin(y) = \frac{1}{2} ln(x) + C$

общее решение

б)

$2y'' - 5y' + 2y = 0 \\ y = {e}^{kx} \\ {e}^{kx} (2 {k}^{2} - 5k + 2) = 0 \\ d = 25 - 16 = 9 \\ k1 = \frac{5 + 3}{4} = 2 \\ k2 = \frac{1}{2} \\ y = C1 {e}^{2x} + C2 {e}^{ \frac{x}{2} }$

общее решение

в)

$y'' - 2y' + y = 0 \\ y = {e}^{kx} \\ {k}^{2} - 2k + 1 = 0 \\ {(k - 1)}^{2} = 0 \\ k1 = k2 = 1 \\ y = C1 {e}^{x} + C2 {e}^{x} x$

общее решение

$y(2) = 1,y'(2) = - 2$

$y' = C1 {e}^{x} + C2 {e}^{x} + C2 {e}^{x} x$

система:

$1 = C1 {e}^{2} + 2C2 {e}^{2} \\ - 2 = C1 {e}^{2} + C2 {e}^{2} + 2C2 {e}^{2} \\ \\ C1 {e}^{2} = 1 - 2C2 {e}^{2} \\ 1 - 2C2 {e}^{2} + C2 {e}^{2} + 2c2 {e}^{2} = - 2 \\ \\ c2 {e}^{2} = - 3 \\ C2 = - \frac{3}{ {e}^{2} } \\ \\ C1 = \frac{1}{ {e}^{2} } (1 - 2C2 {e}^{2} ) = \\ = \frac{1}{ {e}^{2} } (1 - 2 {e}^{2} \times ( - \frac{3}{ {e}^{2} } )) = \\ = \frac{1}{ {e}^{2} } (1 + 6) = \frac{7}{ {e}^{2} }$

частное решение:

$y = \frac{7}{ {e}^{2} } {e}^{x} - \frac{3}{ {e}^{2} } {e}^{x} x = \\ = 7 {e}^{x - 2} - 3 {e}^{x - 2} x$