(t-1)^2y^'+3(t-1)y=t+2

解答

(t - 1)^{2} y^{^{'}} + 3 (t - 1) y = t + 2

y = \frac{t ^{3}}{3 ( t ^{3} - 3 t ^{2} + 3 t - 1 )} + \frac{t ^{2}}{2 ( t ^{3} - 3 t ^{2} + 3 t - 1 )} - \frac{2 t}{t ^{3} - 3 t ^{2} + 3 t - 1} + \frac{c _{1}}{t ^{3} - 3 t ^{2} + 3 t - 1}

求解步骤

(t - 1)^{2} y^{'} (t) + 3 (t - 1) y = t + 2

求解线性方程:

y = \frac{t ^{3}}{3 ( t ^{3} - 3 t ^{2} + 3 t - 1 )} + \frac{t ^{2}}{2 ( t ^{3} - 3 t ^{2} + 3 t - 1 )} - \frac{2 t}{t ^{3} - 3 t ^{2} + 3 t - 1} + \frac{c _{1}}{t ^{3} - 3 t ^{2} + 3 t - 1}

y = \frac{t ^{3}}{3 ( t ^{3} - 3 t ^{2} + 3 t - 1 )} + \frac{t ^{2}}{2 ( t ^{3} - 3 t ^{2} + 3 t - 1 )} - \frac{2 t}{t ^{3} - 3 t ^{2} + 3 t - 1} + \frac{c _{1}}{t ^{3} - 3 t ^{2} + 3 t - 1}

x \to 0 lim (\frac{1 - ln ( x )}{e ^{\frac{1}{x}}})

\frac{d}{d x} (5 x^{5} - x^{3} + 3)

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

\int_{0}^{2} x - x d x

in v erse l a pl a ce \frac{s}{( s ^{2} + 4 s + 5 )}