展开 5+4t-3t^2+3e^-4t-5(sin(4t)+3cos(4t))

解答

展开

5 + 4 t - 3 t^{2} + 3 e^{-} 4 t - 5 (sin (4 t) + 3 cos (4 t))

5 + 4 t - 3 t^{2} + 12 e^{-} t - 5 sin (4 t) - 15 cos (4 t)

求解步骤

5 + 4 t - 3 t^{2} + 3 e^{-} \cdot 4 t - 5 (sin (4 t) + 3 cos (4 t))

= 5 + 4 t - 3 t^{2} + 3 \cdot 4 e^{-} t - 5 (sin (4 t) + 3 cos (4 t))

乘开

- 5 (sin (4 t) + 3 cos (4 t)) : - 5 sin (4 t) - 15 cos (4 t)

= 5 + 4 t - 3 t^{2} + 3 e^{-} \cdot 4 t - 5 sin (4 t) - 15 cos (4 t)

数字相乘:

3 \cdot 4 = 12

= 5 + 4 t - 3 t^{2} + 12 e^{-} t - 5 sin (4 t) - 15 cos (4 t)

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