f^'(x)=4x+9

Lösung

f^{^{'}} (x) = 4 x + 9

f (x) = 2 x^{2} + 9 x + c_{1}

Schritte zur Lösung

f^{'} (x) = 4 x + 9

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

f (x) = \int 4 x + 9 d x

\int 4 x + 9 d x = 2 x^{2} + 9 x + c_{1}

f (x) = 2 x^{2} + 9 x + c_{1}

2 y^{^{'}} + y^{3} = 0

2 x^{2} y^{^{'}} = x^{2} + y^{2}

x^{3} y^{^{'''}} - x^{2} y^{^{'}} = 0

5 x (x + 4 y) y^{^{'}} = 5 y (x - 4 y)

3 y^{^{'}} + 7 y = x^{3} - 2 x^{2} + 1