What is the solution for y^{'''}-2y^{''}+5y^'=-24e^{3x} ?

The solution for y^{'''}-2y^{''}+5y^'=-24e^{3x} is y=c_{1}+e^x(c_{2}cos(2x)+c_{3}sin(2x))-e^{3x}

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y^{'''}-2y^{''}+5y^'=-24e^{3x}

y^{^{'''}} - 2 y^{^{''}} + 5 y^{^{'}} = - 24 e^{3 x}

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

Solution steps

y^{'''} (x) - 2 y^{''} (x) + 5 y^{'} (x) = - 24 e^{3 x}

Solve linear ODE:

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

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

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

y^{^{''}} - 2 y^{^{'}} - 24 y = 6 e^{6 x}

y^{^{''}} + 36 y = 4 sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

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

y^{^{''}} + 8 y^{^{'}} + 16 y = 6 sin (4 t)

What is the solution for y^{'''}-2y^{''}+5y^'=-24e^{3x} ?
The solution for y^{'''}-2y^{''}+5y^'=-24e^{3x} is y=c_{1}+e^x(c_{2}cos(2x)+c_{3}sin(2x))-e^{3x}