What is the solution for y^{''}-y^'-2y=e^xcos(x) ?

The solution for y^{''}-y^'-2y=e^xcos(x) is y=c_{1}e^{2x}+c_{2}e^{-x}+1/10 e^xsin(x)-3/10 e^xcos(x)

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y^{''}-y^'-2y=e^xcos(x)

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = \frac{t}{2} - \frac{t cos ( 2 t )}{2}

y^{^{''}} - 7 y^{^{'}} - 8 y = 8 x^{2} - 2 x

x^{^{''}} + 4 x^{^{'}} + 16 x = 4 sin (2 t)

y^{^{''}} + 16 y = cos (4 t), y (0) = 3, y^{^{'}} (0) = 2

y^{^{''}} - 5 y^{^{'}} - 6 y = 10

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