y^'+0.003y-0.105=0

解答

y^{^{'}} + 0.003 y - 0.105 = 0

y = \frac{e ^{- 0.003 t - c_{1}}}{0.003} + \frac{0.105}{0.003}

求解步骤

y^{'} (t) + 0.003 y - 0.105 = 0

求解可分离方程:

y = \frac{e ^{- 0.003 t - c_{1}}}{0.003} + \frac{0.105}{0.003}

y = \frac{e ^{- 0.003 t - c_{1}}}{0.003} + \frac{0.105}{0.003}

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