What is the solution for y^{''}-2y^'+10y=20x,y(0)=0,y^'(0)=0 ?

The solution for y^{''}-2y^'+10y=20x,y(0)=0,y^'(0)=0 is y=e^x(-2/5 cos(3x)-8/15 sin(3x))+2x+2/5

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y^{''}-2y^'+10y=20x,y(0)=0,y^'(0)=0

y^{^{''}} - 2 y^{^{'}} + 10 y = 20 x, y (0) = 0, y^{^{'}} (0) = 0

y = e^{x} (- \frac{2}{5} cos (3 x) - \frac{8}{15} sin (3 x)) + 2 x + \frac{2}{5}

Solution steps

y^{''} (x) - 2 y^{'} (x) + 10 y = 20 x

Solve linear ODE:

y = e^{x} (- \frac{2}{5} cos (3 x) - \frac{8}{15} sin (3 x)) + 2 x + \frac{2}{5}

y = e^{x} (- \frac{2}{5} cos (3 x) - \frac{8}{15} sin (3 x)) + 2 x + \frac{2}{5}

- a y^{^{''}} + b y - 1 = 0

y^{^{''}} + 16 y = 80 t^{2} - 96 t + 42, y (0) = 0, y^{^{'}} (0) = 14

y^{^{'''}} - 24 y^{^{''}} + 192 y^{^{'}} - 512 y = e^{8 x}

y^{^{'''}} + 18 y^{^{''}} - 864 y = e^{- 12 x}

d (x^{3}) = d x

What is the solution for y^{''}-2y^'+10y=20x,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-2y^'+10y=20x,y(0)=0,y^'(0)=0 is y=e^x(-2/5 cos(3x)-8/15 sin(3x))+2x+2/5