(d-1)^3y=e^x+5^x-1

解答

(d - 1)^{3} y = e^{x} + 5^{x} - 1

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + e^{x} \cdot \int \frac{x ^{2} ( e ^{x} + 5 ^{x} - 1 )}{2 e ^{x}} d x + e^{x} x \cdot \int - \frac{x ( e ^{x} + 5 ^{x} - 1 )}{e ^{x}} d x + \frac{1}{2} e^{x} x^{3} + \frac{5 ^{x} x ^{2}}{2 ( ln ( 5 ) - 1 )} + \frac{x ^{2}}{2}

求解步骤

(d - 1)^{3} y = e^{x} + 5^{x} - 1

求解线性方程:

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + e^{x} \cdot \int \frac{x ^{2} ( e ^{x} + 5 ^{x} - 1 )}{2 e ^{x}} d x + e^{x} x \cdot \int - \frac{x ( e ^{x} + 5 ^{x} - 1 )}{e ^{x}} d x + \frac{1}{2} e^{x} x^{3} + \frac{5 ^{x} x ^{2}}{2 ( ln ( 5 ) - 1 )} + \frac{x ^{2}}{2}

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + e^{x} \cdot \int \frac{x ^{2} ( e ^{x} + 5 ^{x} - 1 )}{2 e ^{x}} d x + e^{x} x \cdot \int - \frac{x ( e ^{x} + 5 ^{x} - 1 )}{e ^{x}} d x + \frac{1}{2} e^{x} x^{3} + \frac{5 ^{x} x ^{2}}{2 ( ln ( 5 ) - 1 )} + \frac{x ^{2}}{2}

y^{^{''}} - 2 y^{^{'}} + y = 12 e^{3 t}, y (0) = - 1, y^{^{'}} (0) = 0

2 y^{^{''}} - 4 y^{^{'}} + 2 y = e^{x} x^{- 1}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = x^{2}

(D^{2} + 2 D + 1) y = e^{- 2 x} + 1

y^{^{''}} + 2 y^{^{'}} + y = x - 2 + sin (x)