What is the solution for y^{''}+2y^'-3y=x+cos(x) ?

The solution for y^{''}+2y^'-3y=x+cos(x) is y=c_{1}e^x+c_{2}e^{-3x}-x/3-2/9+1/10 sin(x)-1/5 cos(x)

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

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

y = c_{1} e^{x} + c_{2} e^{- 3 x} - \frac{x}{3} - \frac{2}{9} + \frac{1}{10} sin (x) - \frac{1}{5} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 3 x} - \frac{x}{3} - \frac{2}{9} + \frac{1}{10} sin (x) - \frac{1}{5} cos (x)

y = c_{1} e^{x} + c_{2} e^{- 3 x} - \frac{x}{3} - \frac{2}{9} + \frac{1}{10} sin (x) - \frac{1}{5} cos (x)

x^{^{''}} + 16 x = 102 e^{- 2 t} cos (4 t)

y^{^{''}} - y = 6 t + 1

y^{^{''}} + 4 y = 3 cos (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y = 6 t + 8

y^{^{''}} + 4 y = sin (t) (t - 2 π)

What is the solution for y^{''}+2y^'-3y=x+cos(x) ?
The solution for y^{''}+2y^'-3y=x+cos(x) is y=c_{1}e^x+c_{2}e^{-3x}-x/3-2/9+1/10 sin(x)-1/5 cos(x)