What is the solution for y^{''}+ay^'+by=x,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+ay^'+by=x,y(0)=0,y^'(0)=0 is

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

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}}

Solution steps

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

Solve linear ODE:

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}}

(D^{2} - 4 D + 3) y = 20 cos (x)

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

y^{^{''}} - 9 y = 20 sin (x) + 3 x^{2}

y^{^{''}} + y^{^{'}} + y = k x

y^{^{''}} - y^{^{'}} - 2 y = 2 x^{3} - 3 x^{2} + 8 x + 1

What is the solution for y^{''}+ay^'+by=x,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+ay^'+by=x,y(0)=0,y^'(0)=0 is