y^{''}+ay^'+by=x,y(0)=0,y^'(0)=0

解

y^{^{''}} + a y^{^{'}} + b y = x, y (0) = 0, y^{^{'}} (0) = 0

y = e^{\frac{x ( - a + a ^{2} - 4 b )}{2}} (\frac{- a ( a ^{2} - 4 b - a ) - 2 b}{2 b ^{2} a ^{2} - 4 b} + \frac{a}{b ^{2}}) - \frac{e ^{\frac{x ( - a - a ^{2} - 4 b )}{2}} ( - a ( a ^{2} - 4 b - a ) - 2 b )}{2 b ^{2} a ^{2} - 4 b} + \frac{x}{b} - \frac{a}{b ^{2}}

解答ステップ

y^{''} (x) + a y^{'} (x) + b y = x

線形 ODE を解く:

y = e^{\frac{x ( - a + a ^{2} - 4 b )}{2}} (\frac{- a ( a ^{2} - 4 b - a ) - 2 b}{2 b ^{2} a ^{2} - 4 b} + \frac{a}{b ^{2}}) - \frac{e ^{\frac{x ( - a - a ^{2} - 4 b )}{2}} ( - a ( a ^{2} - 4 b - a ) - 2 b )}{2 b ^{2} a ^{2} - 4 b} + \frac{x}{b} - \frac{a}{b ^{2}}

y = e^{\frac{x ( - a + a ^{2} - 4 b )}{2}} (\frac{- a ( a ^{2} - 4 b - a ) - 2 b}{2 b ^{2} a ^{2} - 4 b} + \frac{a}{b ^{2}}) - \frac{e ^{\frac{x ( - a - a ^{2} - 4 b )}{2}} ( - a ( a ^{2} - 4 b - a ) - 2 b )}{2 b ^{2} a ^{2} - 4 b} + \frac{x}{b} - \frac{a}{b ^{2}}