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

解答

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

y = (\frac{3}{5} - \frac{7}{5}) e^{2 t} + \frac{7}{5} e^{t}

求解步骤

y^{''} (t) - 3 y^{'} (t) + 2 y = 0

求解线性方程:

y = (\frac{3}{5} - \frac{7}{5}) e^{2 t} + \frac{7}{5} e^{t}

y = (\frac{3}{5} - \frac{7}{5}) e^{2 t} + \frac{7}{5} e^{t}