y^{''}+4y=t^2+6e^t,y(0)=0,y^'(0)=9

解答

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 9

y = - \frac{43}{40} cos (2 t) + \frac{39}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{6}{5} e^{t}

求解步骤

y^{''} (t) + 4 y = t^{2} + 6 e^{t}

求解线性方程:

y = - \frac{43}{40} cos (2 t) + \frac{39}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{6}{5} e^{t}

y = - \frac{43}{40} cos (2 t) + \frac{39}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{6}{5} e^{t}

x^{^{''}} + x = 9 cos (4 t)

y^{^{''}} - 4 y = 3 e^{2 t}

9 y^{^{''}} + 7 y^{^{'}} - y = 24

y^{^{''}} + 16 y = 8 x + 8 sin (x)

y^{^{'''}} + 6 y^{^{''}} + 12 y^{^{'}} + 8 y = 4 e^{- 2 x}