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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 10 y^{^{'}} + 26 y = 4, y (0) = 3, y^{^{'}} (0) = 15

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 5 y = e^{x} cos (2 x)

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 2 e^{- 2 x} + 3

x^{^{''}} - 4 x^{^{'}} + 4 x = e^{2 t}, x (0) = 2, x^{^{'}} (0) = 1

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