y^{''}+3y^'=-48x^2e^3x

解答

y^{^{''}} + 3 y^{^{'}} = - 48 x^{2} e^{3} x

y = c_{1} + c_{2} e^{- 3 x} - 4 e^{3} x^{4} + \frac{16 e ^{3} x ^{3}}{3} - \frac{16 e ^{3} x ^{2}}{3} + \frac{32 e ^{3} x}{9}

求解步骤

y^{''} (x) + 3 y^{'} (x) = - 48 x^{2} e^{3} x

求解线性方程:

y = c_{1} + c_{2} e^{- 3 x} - 4 e^{3} x^{4} + \frac{16 e ^{3} x ^{3}}{3} - \frac{16 e ^{3} x ^{2}}{3} + \frac{32 e ^{3} x}{9}

y = c_{1} + c_{2} e^{- 3 x} - 4 e^{3} x^{4} + \frac{16 e ^{3} x ^{3}}{3} - \frac{16 e ^{3} x ^{2}}{3} + \frac{32 e ^{3} x}{9}

(d^{3} - d^{2} - d + 1) y = 1 + x^{2}

(d^{3} + 4 d) y = 4 cot (2 x)

d (x) = 243 e^{- 0} 0.1 x

(d^{2} - 2 d) y = e^{x} + e^{- x}

(d^{2} - 7 d + 10) y = 5 x + 7