y^{''}-5y^'+6y=5e^{2x}

解答

y^{^{''}} - 5 y^{^{'}} + 6 y = 5 e^{2 x}

y = c_{1} e^{3 x} + c_{2} e^{2 x} - 5 x e^{2 x}

求解步骤

y^{''} (x) - 5 y^{'} (x) + 6 y = 5 e^{2 x}

求解线性方程:

y = c_{1} e^{3 x} + c_{2} e^{2 x} - 5 x e^{2 x}

y = c_{1} e^{3 x} + c_{2} e^{2 x} - 5 x e^{2 x}

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 0 + 6 sin (x)

y^{^{''}} + y = sin (x), y^{^{'}} (\frac{π}{2}) = \frac{3}{2}, y (0) = 0

\frac{d ^{3} y}{d x ^{3}} + \frac{d y}{d x} = tan (x)

3 x^{^{''}} + 25 x^{^{'}} + 48 x = 5 cos (4 t)

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 5