What is the solution for y^{''}-12y^'+35y=96e^t ?

The solution for y^{''}-12y^'+35y=96e^t is y=c_{1}e^{7t}+c_{2}e^{5t}+4e^t

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y^{''}-12y^'+35y=96e^t

y^{^{''}} - 12 y^{^{'}} + 35 y = 96 e^{t}

y = c_{1} e^{7 t} + c_{2} e^{5 t} + 4 e^{t}

Solution steps

y^{''} (t) - 12 y^{'} (t) + 35 y = 96 e^{t}

Solve linear ODE:

y = c_{1} e^{7 t} + c_{2} e^{5 t} + 4 e^{t}

y = c_{1} e^{7 t} + c_{2} e^{5 t} + 4 e^{t}

y^{^{''}} + y + 2 sin (x) = 0

y^{^{''}} - 2 y^{^{'}} + y = 2 sin (t), y^{^{'}} (0) = 2, y (0) = 1

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

y^{^{''}} - 3 y^{^{'}} + y = x

y^{^{''}} + y^{^{'}} = x (e^{3 x} + 1)

What is the solution for y^{''}-12y^'+35y=96e^t ?
The solution for y^{''}-12y^'+35y=96e^t is y=c_{1}e^{7t}+c_{2}e^{5t}+4e^t