y^{''}+6y^'+9=e^{4t},y(0)=0,y^'(0)=0

解答

y^{^{''}} + 6 y^{^{'}} + 9 = e^{4 t}, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{7}{30} - \frac{1}{40} - \frac{7}{30} e^{- 6 t} + \frac{1}{40} e^{4 t} - \frac{3 t}{2}

求解步骤

y^{''} (t) + 6 y^{'} (t) + 9 = e^{4 t}

求解线性方程:

y = \frac{7}{30} - \frac{1}{40} - \frac{7}{30} e^{- 6 t} + \frac{1}{40} e^{4 t} - \frac{3 t}{2}

y = \frac{7}{30} - \frac{1}{40} - \frac{7}{30} e^{- 6 t} + \frac{1}{40} e^{4 t} - \frac{3 t}{2}

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

\int \frac{18 - 12 x}{( 4 x - 1 ) ( x - 4 )} d x

d er i v a t i v e y = \frac{x}{x + 1}

x \to - 9 + lim (\frac{x + 8}{x + 9})

\int \frac{1}{4 x ^{2} + 4 ^{2}} d x