y^{''}-8y^'+20y=10xe^x

解答

y^{^{''}} - 8 y^{^{'}} + 20 y = 10 x e^{x}

y = e^{4 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{10 e ^{x} x}{13} + \frac{60}{169} e^{x}

求解步骤

y^{''} (x) - 8 y^{'} (x) + 20 y = 10 x e^{x}

求解线性方程:

y = e^{4 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{10 e ^{x} x}{13} + \frac{60}{169} e^{x}

y = e^{4 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{10 e ^{x} x}{13} + \frac{60}{169} e^{x}

7 y^{^{''}} + 7 y^{^{'}} + 2 y = 16 + 9 x + x^{2}

y^{^{''}} + y = - 8 x cos^{2} (x)

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 18 x^{2} - 4 x - 15

x^{^{''}} + x^{^{'}} = sin (t)

y^{^{''}} + 4 y^{^{'}} + 4 y = (5 + x) e - 2 x