Question

Помогите, пожалуйста, нужно решить дифференциальные уравнения первого и второго порядка:

Answer 1

1) $y = \int (5x^2 - 3x + 1)\; dx + C = \frac{5x^3}{3} - \frac{3x^2}{2} + x + C$

$y(-1) = 1$

$y(-1) = -\frac{5}{3} - \frac{3}{2} -1 + C = \frac{-10 - 9}{6} - 1 + C =$

$= -\frac{19}{6} - 1 + C = \frac{-19-6}{6} + C = -\frac{25}{6} + C = 1$

$C = 1 + \frac{25}{6} = 1 + 4 + \frac{1}{6} = 5 + \frac{1}{6}$

$y = \frac{5x^3}{3} - \frac{3x^2}{2} + x + 5 + \frac{1}{6}$

2) $\int 6y^3\; dy = \int 9x^2\; dx + C$

$\frac{6y^4}{4} = \frac{9x^3}{3} + C$

$\frac{3y^4}{2} = 3x^3 + C$

$y(3) = 1$

$\frac{3\cdot 1^4}{2} = 3\cdot 3^3 + C$

$\frac{3}{2} = 81 + C$

$C = \frac{3}{2} - 81 = 1{,}5 - 81 = -79{,}5$

$\frac{3y^4}{2} = 3x^3 - 79{,}5$

3) $y' = \int (8x - 6)\; dx = \frac{8x^2}{2} - 6x + C_1$

$y' = 4x^2 - 6x + C_1$

$y'(-1) = -2$

$y'(-1) = 4 + 6 + C_1 = -2$

$10 + C_1 = -2$

$C_1 = -2 -10 = -12$

$y' = 4x^2 - 6x - 12$

$y = \int (4x^2 - 6x - 12)\; dx + C_2 = \frac{4x^3}{3} - \frac{6x^2}{2} - 12x + C_2$

$y = \frac{4x^3}{3} - 3x^2 - 12x + C_2$

$y(-1) = 2$

$y(-1) = -\frac{4}{3} - 3 + 12 + C_2 = 2$

$-1 - \frac{1}{3} + 9 + C_2 = 2$

$C_2 = 2 - 8 + \frac{1}{3} = -6 + \frac{1}{3} = -5 - \frac{2}{3}$

$y = \frac{4x^3}{3} - 3x^2 - 12x - 5 - \frac{2}{3}$