What is the solution for y^{'''}-6y^{''}+25y^'=68e^{2x} ?

The solution for y^{'''}-6y^{''}+25y^'=68e^{2x} is y=c_{1}+e^{3x}(c_{2}cos(4x)+c_{3}sin(4x))+2e^{2x}

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y^{'''}-6y^{''}+25y^'=68e^{2x}

y^{^{'''}} - 6 y^{^{''}} + 25 y^{^{'}} = 68 e^{2 x}

y = c_{1} + e^{3 x} (c_{2} cos (4 x) + c_{3} sin (4 x)) + 2 e^{2 x}

Solution steps

y^{'''} (x) - 6 y^{''} (x) + 25 y^{'} (x) = 68 e^{2 x}

Solve linear ODE:

y = c_{1} + e^{3 x} (c_{2} cos (4 x) + c_{3} sin (4 x)) + 2 e^{2 x}

y = c_{1} + e^{3 x} (c_{2} cos (4 x) + c_{3} sin (4 x)) + 2 e^{2 x}

y^{^{''}} + 4 y^{^{'}} + 13 y = 4 sin (3 t)

(d - 3)^{3} y = e^{3 x}

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

y^{^{''}} + 6 y^{^{'}} + 8 y = x^{2}

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 2 y = e^{4 x}

What is the solution for y^{'''}-6y^{''}+25y^'=68e^{2x} ?
The solution for y^{'''}-6y^{''}+25y^'=68e^{2x} is y=c_{1}+e^{3x}(c_{2}cos(4x)+c_{3}sin(4x))+2e^{2x}