(dy)/(dt)+2y=3t^2+2t-1

解答

\frac{d y}{d t} + 2 y = 3 t^{2} + 2 t - 1

y = \frac{3 t ^{2}}{2} - \frac{t}{2} - \frac{1}{4} + \frac{c _{1}}{e ^{2 t}}

求解步骤

\frac{d y}{d t} + 2 y = 3 t^{2} + 2 t - 1

求解线性方程:

y = \frac{3 t ^{2}}{2} - \frac{t}{2} - \frac{1}{4} + \frac{c _{1}}{e ^{2 t}}

y = \frac{3 t ^{2}}{2} - \frac{t}{2} - \frac{1}{4} + \frac{c _{1}}{e ^{2 t}}

n = 1 \sum \infty \frac{8 ^{n}}{n ^{4}}

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

\int \frac{x}{16 - x ^{4}} d x

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

\int_{4}^{9} \frac{t - 3}{t} d t