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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{2} e^{3 t}

y^{^{''}} - 25 y = cosh (5 x)

y^{^{''}} - 3 y^{^{'}} - 4 y = e^{- x}

3 y^{^{''}} + 27 y = 2 cos (3 t)

y^{^{''}} + 16 y = 32 t^{2} - 48 t + 100, y (0) = 0, y^{^{'}} (0) = 1

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