What is the solution for y^{''}-5y^'+6y=2e^t ?

The solution for y^{''}-5y^'+6y=2e^t is y=c_{1}e^{3t}+c_{2}e^{2t}+e^t

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y^{''}-5y^'+6y=2e^t

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{t}

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

Solution steps

y^{''} (t) - 5 y^{'} (t) + 6 y = 2 e^{t}

Solve linear ODE:

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

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

(5, 2), (2, - 3)

\frac{d}{d x} (\frac{x}{x ^{4} + 1})

y = (x^{3} - 16 x)^{14}, (4, 0)

\frac{1}{3 ( 5 - x ^{3} ) ^{5}}

y^{^{'}} + 4 x y = x^{3} e^{x^{2}}, y (0) = - 1

What is the solution for y^{''}-5y^'+6y=2e^t ?
The solution for y^{''}-5y^'+6y=2e^t is y=c_{1}e^{3t}+c_{2}e^{2t}+e^t