What is the solution for 2y^{''}+3y^'-2y=-8x-13,y(0)=0,y^'(0)=0 ?

The solution for 2y^{''}+3y^'-2y=-8x-13,y(0)=0,y^'(0)=0 is y=(9/10-25/2)e^{x/2}-9/10 e^{-2x}+4x+25/2

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2y^{''}+3y^'-2y=-8x-13,y(0)=0,y^'(0)=0

2 y^{^{''}} + 3 y^{^{'}} - 2 y = - 8 x - 13, y (0) = 0, y^{^{'}} (0) = 0

y = (\frac{9}{10} - \frac{25}{2}) e^{\frac{x}{2}} - \frac{9}{10} e^{- 2 x} + 4 x + \frac{25}{2}

Solution steps

2 y^{''} (x) + 3 y^{'} (x) - 2 y = - 8 x - 13

Solve linear ODE:

y = (\frac{9}{10} - \frac{25}{2}) e^{\frac{x}{2}} - \frac{9}{10} e^{- 2 x} + 4 x + \frac{25}{2}

y = (\frac{9}{10} - \frac{25}{2}) e^{\frac{x}{2}} - \frac{9}{10} e^{- 2 x} + 4 x + \frac{25}{2}

2 x^{^{''}} - x^{^{'}} + 4 x = cos (3 t)

y^{^{''}} + 9 y = 5 sin (x)

y^{^{''}} + 7 y^{^{'}} + 10 y = x e^{- 2 x}

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 e^{- t}

What is the solution for 2y^{''}+3y^'-2y=-8x-13,y(0)=0,y^'(0)=0 ?
The solution for 2y^{''}+3y^'-2y=-8x-13,y(0)=0,y^'(0)=0 is y=(9/10-25/2)e^{x/2}-9/10 e^{-2x}+4x+25/2