y^{''}+16=0

解答

y^{^{''}} + 16 = 0

y = - 8 t^{2} + c_{1} t + c_{2}

求解步骤

y^{''} (t) + 16 = 0

整理

y^{''} (t) : y^{''} (t) = - 16

解

y^{''} (t) = - 16 : y^{'} (t) = - 16 t + c_{1}

解

y^{'} (t) = - 16 t + c_{1} : y = - 8 t^{2} + c_{1} t + c_{2}

y = - 8 t^{2} + c_{1} t + c_{2}

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

\int \frac{x}{4} d x

x \to 8 lim (\frac{∣ x - 8 ∣}{x - 8})

x \to 6 + lim (\frac{2 + x}{6 - x})

d er i v a t i v e f (x) = 3 x^{2 x + 3} + e^{2}