y^{''}-5y^'+6y=0,y(1)=e^2,y^'(1)=3e^2

解答

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (1) = e^{2}, y^{^{'}} (1) = 3 e^{2}

y = e^{3 t - 1}

求解步骤

y^{''} (t) - 5 y^{'} (t) + 6 y = 0

求解线性方程:

y = e^{3 t - 1}

y = e^{3 t - 1}

\frac{d}{d x} (\frac{1}{4} e^{- \frac{x}{2}})

x \to π lim (sin (x - π))

x \to 0 + lim (- \frac{3}{x})

y^{^{'}} = \frac{4 x ^{2} + y ^{2}}{x y}

\frac{\partial}{\partial x} (3 x + y)