What is the solution for y^{''}-11y^'+24y=280e^t ?

The solution for y^{''}-11y^'+24y=280e^t is y=c_{1}e^{8t}+c_{2}e^{3t}+20e^t

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y^{''}-11y^'+24y=280e^t

y^{^{''}} - 11 y^{^{'}} + 24 y = 280 e^{t}

y = c_{1} e^{8 t} + c_{2} e^{3 t} + 20 e^{t}

Solution steps

y^{''} (t) - 11 y^{'} (t) + 24 y = 280 e^{t}

Solve linear ODE:

y = c_{1} e^{8 t} + c_{2} e^{3 t} + 20 e^{t}

y = c_{1} e^{8 t} + c_{2} e^{3 t} + 20 e^{t}

y^{^{''}} + 9 y = 5 sec (3 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = \frac{e ^{3 x}}{e ^{2 x} + 1}

y^{^{''}} + 10000 y = 100 sin (50 x)

y^{^{''}} + 10 y^{^{'}} + 15 = 0

\frac{d ^{2} x}{d t ^{2}} + \frac{d x}{d t} + x = 1

What is the solution for y^{''}-11y^'+24y=280e^t ?
The solution for y^{''}-11y^'+24y=280e^t is y=c_{1}e^{8t}+c_{2}e^{3t}+20e^t