Решите диф.ур.
[tex]y`+\frac{2}{3}xy=\frac{2}{3}xe^{-2x^{2} } y^{-2}, y(0)=-1[/tex]

[tex] y' + \frac{2}{3}\cdot x y = \frac{2}{3}\cdot x e^{-2x^2}y^{-2} [/tex]

[tex] y(0) = -1 [/tex]

[tex] y^2\cdot y' + \frac{2}{3}\cdot x y^3 = \frac{2}{3}\cdot x e^{-2x^2} [/tex]

[tex] y^2\cdot y' \equiv \frac{(y^3)'}{3} [/tex]

[tex] \frac{1}{3}\cdot (y^3)' + \frac{2}{3}\cdot x y^3 = \frac{2}{3}\cdot x e^{-2x^2} [/tex]

[tex] (y^3)' + 2x y^3 = 2x\cdot e^{-2x^2} [/tex]

[tex] y^3 = u = u(x) [/tex]

[tex] u' + 2xu = 2x\cdot e^{-2x^2} [/tex]

[tex] A = A(x) [/tex]

[tex] A\cdot u' + 2Axu = (A\cdot u)' = A\cdot u' + A'\cdot u [/tex]

[tex] 2Ax = A'  = \frac{dA}{dx} [/tex]

[tex] 2x = \frac{1}{A}\cdot\frac{dA}{dx} [/tex]

[tex] \int 2x\; dx = \int \frac{1}{A}\cdot\frac{dA}{dx}\; dx + C [/tex]

[tex] x^2 = \int \frac{1}{A}\; dA + C [/tex]

[tex] x^2 = \ln|A| + C [/tex]

[tex] \ln|A| = x^2 - C [/tex]

[tex] |A| = e^{x^2 - C} = e^{-C}\cdot e^{x^2} [/tex]

[tex] A = \pm e^{-C}\cdot e^{x^2} [/tex]

[tex] A = C_1\cdot e^{x^2} [/tex]

Положим C_1  = 1.

[tex] A = e^{x^2} [/tex]

[tex] e^{x^2}u' + 2x\cdot e^{x^2}u = 2x\cdot e^{-2x^2}\cdot e^{x^2} [/tex]

[tex] \left( e^{x^2} u\right)' = 2x\cdot e^{-x^2} [/tex]

[tex] e^{x^2} u = \int 2x\cdot e^{-x^2}\; dx + C = [/tex]

[tex] = \int e^{-x^2}\; d(x^2) + C = - \int e^{-x^2}\; d(-x^2) + C = [/tex]

[tex] = -e^{-x^2} + C [/tex]

[tex] e^{x^2} u = C - e^{-x^2} [/tex]

[tex] u = C\cdot e^{-x^2} - e^{-x^2}\cdot e^{-x^2} = C\cdot e^{-x^2} - e^{-2x^2} [/tex]

[tex] y^3 = C\cdot e^{-x^2} - e^{-2x^2} [/tex]

[tex] y = \sqrt[3]{C\cdot e^{-x^2} - e^{-2x^2} } [/tex]

[tex] y(0) = \sqrt[3]{C\cdot e^0 - e^0 } = \sqrt[3]{ C - 1 } = -1 [/tex]

[tex] C - 1 = (-1)^3 = -1 [/tex]

[tex] C = 0 [/tex]

[tex] y = \sqrt[3]{ -e^{-2x^2} } = - e^{-\frac{2x^2}{3}} [/tex]