What is the solution for y^{''}-10y^'+24y=150e^t ?

The solution for y^{''}-10y^'+24y=150e^t is y=c_{1}e^{6t}+c_{2}e^{4t}+10e^t

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y^{''}-10y^'+24y=150e^t

y^{^{''}} - 10 y^{^{'}} + 24 y = 150 e^{t}

y = c_{1} e^{6 t} + c_{2} e^{4 t} + 10 e^{t}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 24 y = 150 e^{t}

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{4 t} + 10 e^{t}

y = c_{1} e^{6 t} + c_{2} e^{4 t} + 10 e^{t}

y^{^{''}} - 2 y^{^{'}} = 12 x + 10

y^{^{''}} + 2 y^{^{'}} + 17 y = 20 e^{t}, y (0) = 2, y^{^{''}} (0) = - 4

y^{^{''}} + 7 y^{^{'}} + 12 y = 2 e^{- t}

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

2 x^{^{''}} + 3 x^{^{'}} + 5 x = 2 t

What is the solution for y^{''}-10y^'+24y=150e^t ?
The solution for y^{''}-10y^'+24y=150e^t is y=c_{1}e^{6t}+c_{2}e^{4t}+10e^t