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

Lösung

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

y = \frac{- 3 + 4 x ^{3} - 4 x + 25}{2}

Schritte zur Lösung

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

Solve separable ODE:

y = \frac{- 3 + 4 x ^{3} - 4 x + 25}{2}

y = \frac{- 3 + 4 x ^{3} - 4 x + 25}{2}

\frac{d}{d x} (sin (y)) = x

y^{^{'}} - y = y^{3}

d y = (y x^{2} - 1.1 y) d x

\frac{d y}{d x} = 6 - 3 y

3 y^{^{'}} = \frac{4 x}{y ^{2}}