What is the solution for y^{''}+y-5=e^{2x}-3+cos(3x) ?

The solution for y^{''}+y-5=e^{2x}-3+cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)+1/5 e^{2x}-1/8 cos(3x)+2

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y^{''}+y-5=e^{2x}-3+cos(3x)

y^{^{''}} + y - 5 = e^{2 x} - 3 + cos (3 x)

y = c_{1} cos (x) + c_{2} sin (x) + \frac{1}{5} e^{2 x} - \frac{1}{8} cos (3 x) + 2

Solution steps

y^{''} (x) + y - 5 = e^{2 x} - 3 + cos (3 x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + \frac{1}{5} e^{2 x} - \frac{1}{8} cos (3 x) + 2

y = c_{1} cos (x) + c_{2} sin (x) + \frac{1}{5} e^{2 x} - \frac{1}{8} cos (3 x) + 2

y^{^{''}} - 4 y = 2 e^{3 x}

y^{^{''}} + 5 y^{^{'}} = - 6 e^{- 5 x}

y^{^{''}} + 2 y^{^{'}} + 4 y = 3 sin (5 t)

y^{^{''}} - 4 y^{^{'}} - 12 y = 3 cos (x)

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

What is the solution for y^{''}+y-5=e^{2x}-3+cos(3x) ?
The solution for y^{''}+y-5=e^{2x}-3+cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)+1/5 e^{2x}-1/8 cos(3x)+2