What is the solution for y^{''}+2y^'-3y=7cos(3x) ?

The solution for y^{''}+2y^'-3y=7cos(3x) is y=c_{1}e^x+c_{2}e^{-3x}+7/30 sin(3x)-7/15 cos(3x)

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y^{''}+2y^'-3y=7cos(3x)

y^{^{''}} + 2 y^{^{'}} - 3 y = 7 cos (3 x)

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{7}{30} sin (3 x) - \frac{7}{15} cos (3 x)

Solution steps

y^{''} (x) + 2 y^{'} (x) - 3 y = 7 cos (3 x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{7}{30} sin (3 x) - \frac{7}{15} cos (3 x)

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{7}{30} sin (3 x) - \frac{7}{15} cos (3 x)

y^{^{''}} - 9 y^{^{'}} + 4 y = x e^{x}

y^{^{''}} + 4 y = sec^{2} (x)

y^{^{''}} + 2 y^{^{'}} - 3 y = 1, y (0) = e^{x} - \frac{1}{3}

y^{^{'''}} - 12 y^{^{''}} + 256 y = e^{8 x}

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

What is the solution for y^{''}+2y^'-3y=7cos(3x) ?
The solution for y^{''}+2y^'-3y=7cos(3x) is y=c_{1}e^x+c_{2}e^{-3x}+7/30 sin(3x)-7/15 cos(3x)