What is the solution for y^{''}-2y^'+17y=e^xsin(x) ?

The solution for y^{''}-2y^'+17y=e^xsin(x) is y=e^x(c_{1}cos(4x)+c_{2}sin(4x))+1/15 e^xsin(x)

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y^{''}-2y^'+17y=e^xsin(x)

y^{^{''}} - 2 y^{^{'}} + 17 y = e^{x} sin (x)

y = e^{x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + \frac{1}{15} e^{x} sin (x)

Solution steps

y^{''} (x) - 2 y^{'} (x) + 17 y = e^{x} sin (x)

Solve linear ODE:

y = e^{x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + \frac{1}{15} e^{x} sin (x)

y = e^{x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + \frac{1}{15} e^{x} sin (x)

y^{^{''}} - 4 y^{^{'}} - 5 y = 2 + e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t^{2} - 5

4 x^{^{''}} + x = t^{2} + 2 cos (3 t)

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

y^{^{''}} + y = cos (2 t), y (0) = 0, y^{^{'}} (0) = 1

What is the solution for y^{''}-2y^'+17y=e^xsin(x) ?
The solution for y^{''}-2y^'+17y=e^xsin(x) is y=e^x(c_{1}cos(4x)+c_{2}sin(4x))+1/15 e^xsin(x)