x(3y^2-2)y^'(x)-y^3=0,y(1)=2

Lösung

x (3 y^{2} - 2) y^{^{'}} (x) - y^{3} = 0, y (1) = 2

3 x y' (x) y^{2} - 2 x y' (x) - 16 \circ

Schritte zur Lösung

Finde

x (3 y^{2} - 2) y^{'} (x) - y^{3} \circ y

gegeben

x (3 y^{2} - 2) y^{'} (x) - y^{3} = 0, y (1) = 2

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

y^{2} : 2^{2}

y^{3} : 2^{3}

y : 2

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

Multipliziere aus

x (3 y^{2} - 2) y^{'} (x) - 2^{3} \circ 2 : 3 x y^{'} (x) y^{2} - 2 x y^{'} (x) - 16 \circ

= 3 x y^{'} (x) y^{2} - 2 x y^{'} (x) - 16 \circ

y^{2} d y = x d x

tan^{2} (x) cos^{2} (y) d x = d y

\frac{d y}{d x} = \frac{- 6}{4} x^{5} y^{8}

\frac{d y}{d x} = \frac{2 y}{x + 3}

x \frac{d y}{d x} = \frac{1 - 4 y ^{2}}{3 y}