y^{''}-4y^'+4y=(3x+8)e^{2x}+8x^2

解答

y^{^{''}} - 4 y^{^{'}} + 4 y = (3 x + 8) e^{2 x} + 8 x^{2}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{e ^{2 x} x ^{3}}{2} + 4 x^{2} e^{2 x} + 2 x^{2} + 4 x + 3

求解步骤

y^{''} (x) - 4 y^{'} (x) + 4 y = (3 x + 8) e^{2 x} + 8 x^{2}

求解线性方程:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{e ^{2 x} x ^{3}}{2} + 4 x^{2} e^{2 x} + 2 x^{2} + 4 x + 3

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{e ^{2 x} x ^{3}}{2} + 4 x^{2} e^{2 x} + 2 x^{2} + 4 x + 3

\int y ln (y) - \frac{1}{9 - y ^{2}} d y

\frac{\partial}{\partial x} (2 x - 10)

\int k t d t

\int x^{3} z e^{x z} d x

n = 0 \sum \infty - 6 (- \frac{1}{3})^{2 n}