-4y^'+12x^2=0

Lösung

- 4 y^{^{'}} + 12 x^{2} = 0

y = x^{3} + c_{1}

Schritte zur Lösung

- 4 y^{'} (x) + 12 x^{2} = 0

Isoliere

y^{'} (x) : y^{'} (x) = 3 x^{2}

Wenn

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

dann

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

y = \int 3 x^{2} d x

\int 3 x^{2} d x = x^{3} + c_{1}

y = x^{3} + c_{1}

\frac{\partial}{\partial x} (3 y^{2} cos (5 x))

\frac{\partial}{\partial x} (2 x y e^{x y^{2}} - 3 y^{2})

\int \frac{1}{( ln ( x ) ) ^{2}} d x

\frac{\partial}{\partial y} (2 arctan (\frac{y}{x}))

t an g e n t f (x) = 2 + 3 x - 5 x^{2}, at x = 1