What is the solution for y^{''}-2y^'+26y=e^xcos(5x) ?

The solution for y^{''}-2y^'+26y=e^xcos(5x) is y=e^x(c_{1}cos(5x)+c_{2}sin(5x))+(e^xxsin(5x))/(10)

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y^{''}-2y^'+26y=e^xcos(5x)

y^{^{''}} - 2 y^{^{'}} + 26 y = e^{x} cos (5 x)

y = e^{x} (c_{1} cos (5 x) + c_{2} sin (5 x)) + \frac{e ^{x} x sin ( 5 x )}{10}

Solution steps

y^{''} (x) - 2 y^{'} (x) + 26 y = e^{x} cos (5 x)

Solve linear ODE:

y = e^{x} (c_{1} cos (5 x) + c_{2} sin (5 x)) + \frac{e ^{x} x sin ( 5 x )}{10}

y = e^{x} (c_{1} cos (5 x) + c_{2} sin (5 x)) + \frac{e ^{x} x sin ( 5 x )}{10}

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

x^{^{''}} + 5 x^{^{'}} + 6 x = 12 e^{- 5 t} + 12 e^{- t} + 24

y^{^{''}} + 6 y^{^{'}} + 5 y = 20

y^{^{''}} + y = e^{3 t} - 4 t cos (5 t)

y^{^{''''}} + 8 y^{^{''}} + 16 y = cos (x)

What is the solution for y^{''}-2y^'+26y=e^xcos(5x) ?
The solution for y^{''}-2y^'+26y=e^xcos(5x) is y=e^x(c_{1}cos(5x)+c_{2}sin(5x))+(e^xxsin(5x))/(10)