What is the solution for y^{''}-2y^'+5y=10sin(x) ?

The solution for y^{''}-2y^'+5y=10sin(x) is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+2sin(x)+cos(x)

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y^{''}-2y^'+5y=10sin(x)

y^{^{''}} - 2 y^{^{'}} + 5 y = 10 sin (x)

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 2 sin (x) + cos (x)

Solution steps

y^{''} (x) - 2 y^{'} (x) + 5 y = 10 sin (x)

Solve linear ODE:

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 2 sin (x) + cos (x)

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 2 sin (x) + cos (x)

y^{^{''}} + \frac{1}{8} y^{^{'}} + y = 3 cos (x)

y^{^{''}} + 8 y^{^{'}} + 16 = 24 x + 2

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

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

y^{^{''}} + 36 y = 6 tan (6 t) - 2 e^{3 t}

What is the solution for y^{''}-2y^'+5y=10sin(x) ?
The solution for y^{''}-2y^'+5y=10sin(x) is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+2sin(x)+cos(x)