Question

2y'=y^3/x^2 при условии y(-1)=1

Answer 1

$x=\frac{y^{2}}{Cy^{2}+1}, \quad C-const; \quad x=\frac{y^{2}}{1-2y^{2}};$

Пошаговое объяснение:

$2y'=\frac{y^{3}}{x^{2}};$

$2\frac{dy}{dx}=\frac{y^{3}}{x^{2}};$

$2\frac{dy}{y^{3}dx}=\frac{1}{x^{2}};$

$2\frac{dy}{y^{3}}=\frac{dx}{x^{2}};$

$2\int\limits {\frac{1}{y^{3}}} \, dy =\int\limits {\frac{1}{x^{2}}} \, dx;$

$2\int\limits {y^{-3}} \, dy =\int\limits {x^{-2}} \, dx;$

$2 \cdot \frac{y^{-3+1}}{-3+1}=\frac{x^{-2+1}}{-2+1}+C, \quad C-const;$

$2 \cdot \frac{y^{-2}}{-2}=\frac{x^{-1}}{-1}+C, \quad C-const;$

$-y^{-2}=-x^{-1}+C, \quad C-const;$

$y^{-2}=x^{-1}-C, \quad C-const;$

$\frac{1}{x}-C=\frac{1}{y^{2}}, \quad C-const;$

$\frac{1}{x}=C+\frac{1}{y^{2}}, \quad C-const;$

$\frac{1}{x}=\frac{Cy^{2}+1}{y^{2}}, \quad C-const;$

$\frac{x}{1}=\frac{y^{2}}{Cy^{2}+1}, \quad C-const;$

$x=\frac{y^{2}}{Cy^{2}+1}, \quad C-const;$

$y(-1)=1 \Rightarrow x=-1, \quad y=1;$

$-1=\frac{1^{2}}{C \cdot 1^{2}+1};$

$-1=\frac{1}{C+1};$

$C+1=\frac{1}{-1};$

$C+1=-1;$

$C=-1-1;$

$C=-2;$

$x=\frac{y^{2}}{1-2y^{2}};$