derivative of y^3=50x

Lösung

\frac{d}{d x} (y^{3}) = 50 x

y = 3 25 x^{2} + c_{1}, y = - \frac{3 25 x ^{2} + c _{1}}{2} + \frac{3 25 x ^{2} + c _{1} 3}{2} i, y = - \frac{3 25 x ^{2} + c _{1}}{2} - \frac{3 25 x ^{2} + c _{1} 3}{2} i

Schritte zur Lösung

\frac{d}{d x} (y^{3}) = 50 x

Wenn

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

dann

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

y^{3} = \int 50 x d x

\int 50 x d x = 25 x^{2} + c_{1}

y^{3} = 25 x^{2} + c_{1}

Isoliere

y : y = 3 25 x^{2} + c_{1}, y = - \frac{3 25 x ^{2} + c _{1}}{2} + \frac{3 25 x ^{2} + c _{1} 3}{2} i, y = - \frac{3 25 x ^{2} + c _{1}}{2} - \frac{3 25 x ^{2} + c _{1} 3}{2} i

y = 3 25 x^{2} + c_{1}, y = - \frac{3 25 x ^{2} + c _{1}}{2} + \frac{3 25 x ^{2} + c _{1} 3}{2} i, y = - \frac{3 25 x ^{2} + c _{1}}{2} - \frac{3 25 x ^{2} + c _{1} 3}{2} i

\frac{d y}{d x} = \frac{1}{y - 2}

\frac{\frac{d y}{d t}}{y} = \frac{( 2 t + 1 )}{t ^{2}}

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

\frac{d x}{d t} = x^{2} + \frac{1}{9}

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