y^{''}-4y^'+4y=4-8t-(18te^{2t}+2e^{2t})

解答

y^{^{''}} - 4 y^{^{'}} + 4 y = 4 - 8 t - (18 t e^{2 t} + 2 e^{2 t})

y = c_{1} e^{2 t} + c_{2} t e^{2 t} - 2 t - 1 - 3 e^{2 t} t^{3} - e^{2 t} t^{2}

求解步骤

y^{''} (t) - 4 y^{'} (t) + 4 y = 4 - 8 t - (18 t e^{2 t} + 2 e^{2 t})

求解线性方程:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} - 2 t - 1 - 3 e^{2 t} t^{3} - e^{2 t} t^{2}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} - 2 t - 1 - 3 e^{2 t} t^{3} - e^{2 t} t^{2}

\int 2 \cdot x \cdot ln (x) d x

\frac{\partial}{\partial y} (- \frac{1}{2} x - 3 y + 1)

\int u cos (u) d u

sin (x) sin (y) d x + cos (x) cos (y) d y = 0

\frac{\partial}{\partial y} (x^{2} e^{2 y})