What is the solution for y^{'''}-2y^{''}+10y^'=250e^{5x} ?

The solution for y^{'''}-2y^{''}+10y^'=250e^{5x} is y=c_{1}+e^x(c_{2}cos(3x)+c_{3}sin(3x))+2e^{5x}

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y^{'''}-2y^{''}+10y^'=250e^{5x}

y^{^{'''}} - 2 y^{^{''}} + 10 y^{^{'}} = 250 e^{5 x}

y = c_{1} + e^{x} (c_{2} cos (3 x) + c_{3} sin (3 x)) + 2 e^{5 x}

Solution steps

y^{'''} (x) - 2 y^{''} (x) + 10 y^{'} (x) = 250 e^{5 x}

Solve linear ODE:

y = c_{1} + e^{x} (c_{2} cos (3 x) + c_{3} sin (3 x)) + 2 e^{5 x}

y = c_{1} + e^{x} (c_{2} cos (3 x) + c_{3} sin (3 x)) + 2 e^{5 x}

y^{^{'''}} - y^{^{''}} - 4 y^{^{'}} + 4 = 0

y^{^{''}} - 9 y = 4 e^{- 5 t}

y^{^{''}} + 4 y = 3 cos (a t)

y^{^{'''}} + 9 y^{^{''}} + 27 y^{^{'}} + 27 y = 7 e^{- 3 x}

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} = 2 x^{2} - x + 1

What is the solution for y^{'''}-2y^{''}+10y^'=250e^{5x} ?
The solution for y^{'''}-2y^{''}+10y^'=250e^{5x} is y=c_{1}+e^x(c_{2}cos(3x)+c_{3}sin(3x))+2e^{5x}