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Ответ:

1)

[tex] \frac{dy}{dx} = \frac{x}{y} [/tex]

[tex] \frac{dy}{dx} = \frac{x}{y} \\ ydy = xdx \\ \int ydy = \int xdx[/tex]

[tex] \frac{ {y}^{2} }{2} = \frac{ {x}^{2} }{2} + c[/tex]

Подставим значения y(1)=-2, значит при x=1, y=-2.

[tex] \displaystyle \frac{ {( - 2)}^{2} }{2} = \frac{ {1}^{2} }{2} + c \\ \frac{4}{2} = \frac{1}{2} + c \\ c = \frac{4}{2} - \frac{1}{2} \\ c = \frac{3}{2} \\ c = 1.5[/tex]

[tex] \frac{ {y}^{2} }{2} = \frac{ {x}^{2} }{2} + 1.5 \\ {y}^{2} = 2( \frac{ {x}^{2} }{2} + 1.5) \\ {y}^{2} = {x}^{2} + 3 \\ y = \sqrt{ {x}^{2} + 3 } [/tex]

2)

[tex] \frac{dy}{dx} = 3y {x}^{2} \\ \frac{dy}{y} = 3 {x}^{2} dx [/tex]

[tex] \int \frac{dy}{y} = \int 3 {x}^{2} dx[/tex]

[tex] ln|y| = 3 \times \frac{ {x}^{3} }{3} + c \\ ln |y| = {x}^{3} + c[/tex]

При x=1, y=1

[tex] ln(1) = {1}^{3} + c \\ 0 = 1 + c \\ c = - 1[/tex]

[tex] ln(y) = {x}^{3} - 1 \\ y = {e}^{ {x}^{3} - 1 } [/tex]