derivative of 2y^2=3-x

Lösung

\frac{d}{d x} (2 y^{2}) = 3 - x

y = \frac{6 x + c _{1} - x ^{2}}{2}, y = - \frac{6 x + c _{1} - x ^{2}}{2}

Schritte zur Lösung

\frac{d}{d x} (2 y^{2}) = 3 - x

Wenn

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

dann

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

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

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

2 y^{2} = 3 x - \frac{x ^{2}}{2} + c_{1}

Isoliere

y : y = \frac{6 x + 2 c _{1} - x ^{2}}{2}, y = - \frac{6 x + 2 c _{1} - x ^{2}}{2}

y = \frac{6 x + 2 c _{1} - x ^{2}}{2}, y = - \frac{6 x + 2 c _{1} - x ^{2}}{2}

Vereinfache

y = \frac{6 x + c _{1} - x ^{2}}{2}, y = - \frac{6 x + c _{1} - x ^{2}}{2}

y^{^{'}} = e^{y} x^{6}

\frac{d y}{d x} = 64 y x^{15}

2 x y y^{^{'}} - y^{2} + 1 = 0

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

d y + 0.15 (y - 25) d x = 0