y^'=-2x(y^2-3y+2),y(0)=3

Lösung

y^{^{'}} = - 2 x (y^{2} - 3 y + 2), y (0) = 3

y = - \frac{e ^{- x^{2}}}{2 - e ^{- x^{2}}} + \frac{4}{2 - e ^{- x^{2}}}

Schritte zur Lösung

y^{'} (x) = - 2 x (y^{2} - 3 y + 2)

Solve separable ODE:

y = - \frac{e ^{- x^{2}}}{2 - e ^{- x^{2}}} + \frac{4}{2 - e ^{- x^{2}}}

y = - \frac{e ^{- x^{2}}}{2 - e ^{- x^{2}}} + \frac{4}{2 - e ^{- x^{2}}}

y^{^{'}} = \frac{3 y}{x + 6}

x^{^{'}} = - x + 1

4 y x \frac{d x}{d y} = x^{2} + 1

(x^{2} - 3) y^{^{'}} = x (y^{2} + 16), y (2) = 4

\frac{d y}{d x} = 3 y^{2} - y^{2} sin (x)