Question

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

Answer 1

$y' + \frac{2}{3}\cdot x y = \frac{2}{3}\cdot x e^{-2x^2}y^{-2}$

$y(0) = -1$

$y^2\cdot y' + \frac{2}{3}\cdot x y^3 = \frac{2}{3}\cdot x e^{-2x^2}$

$y^2\cdot y' \equiv \frac{(y^3)'}{3}$

$\frac{1}{3}\cdot (y^3)' + \frac{2}{3}\cdot x y^3 = \frac{2}{3}\cdot x e^{-2x^2}$

$(y^3)' + 2x y^3 = 2x\cdot e^{-2x^2}$

$y^3 = u = u(x)$

$u' + 2xu = 2x\cdot e^{-2x^2}$

$A = A(x)$

$A\cdot u' + 2Axu = (A\cdot u)' = A\cdot u' + A'\cdot u$

$2Ax = A'  = \frac{dA}{dx}$

$2x = \frac{1}{A}\cdot\frac{dA}{dx}$

$\int 2x\; dx = \int \frac{1}{A}\cdot\frac{dA}{dx}\; dx + C$

$x^2 = \int \frac{1}{A}\; dA + C$

$x^2 = \ln|A| + C$

$\ln|A| = x^2 - C$

$|A| = e^{x^2 - C} = e^{-C}\cdot e^{x^2}$

$A = \pm e^{-C}\cdot e^{x^2}$

$A = C_1\cdot e^{x^2}$

Положим C_1  = 1.

$A = e^{x^2}$

$e^{x^2}u' + 2x\cdot e^{x^2}u = 2x\cdot e^{-2x^2}\cdot e^{x^2}$

$\left( e^{x^2} u\right)' = 2x\cdot e^{-x^2}$

$e^{x^2} u = \int 2x\cdot e^{-x^2}\; dx + C =$

$= \int e^{-x^2}\; d(x^2) + C = - \int e^{-x^2}\; d(-x^2) + C =$

$= -e^{-x^2} + C$

$e^{x^2} u = C - e^{-x^2}$

$u = C\cdot e^{-x^2} - e^{-x^2}\cdot e^{-x^2} = C\cdot e^{-x^2} - e^{-2x^2}$

$y^3 = C\cdot e^{-x^2} - e^{-2x^2}$

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

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

$C - 1 = (-1)^3 = -1$

$C = 0$

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