2y^{''}+y^'-y=3x+4,y(0)=1,y^'(0)=1

解答

2 y^{^{''}} + y^{^{'}} - y = 3 x + 4, y (0) = 1, y^{^{'}} (0) = 1

y = - 3 x + 8 e^{\frac{x}{2}} - 7

求解步骤

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

求解线性方程:

y = - 3 x + 8 e^{\frac{x}{2}} - 7

y = - 3 x + 8 e^{\frac{x}{2}} - 7

y^{^{''}} - y^{^{'}} - 6 y = 3 e^{2 x}

y^{^{''}} + 7 y^{^{'}} - 12 y = e^{3} x - 2 e^{x} - 5 x

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

\frac{d ^{2} y}{d x ^{2}} + 4 y = cos (\frac{9}{4} x)

\frac{d ^{2} y}{d x ^{2}} + a y^{^{'}} = b