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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 12 y^{^{'}} + 36 y = 12 e^{6 x}

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

y^{^{''}} + 3 y^{^{'}} + 2 y = t H (t - 3), y (0) = 0, y^{^{'}} (0) = 1

4 \cdot y^{^{''}} - 11 \cdot y^{^{'}} - 3 y = 5 \cdot e^{3 x}

\frac{d ^{2} x}{d t ^{2}} + x = 6 cos (2 t), x (0) = 3

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