What is the solution for y^{''}-11y^'+18y=6 ?

The solution for y^{''}-11y^'+18y=6 is y=c_{1}e^{9t}+c_{2}e^{2t}+1/3

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y^{''}-11y^'+18y=6

y^{^{''}} - 11 y^{^{'}} + 18 y = 6

y = c_{1} e^{9 t} + c_{2} e^{2 t} + \frac{1}{3}

Solution steps

y^{''} (t) - 11 y^{'} (t) + 18 y = 6

Solve linear ODE:

y = c_{1} e^{9 t} + c_{2} e^{2 t} + \frac{1}{3}

y = c_{1} e^{9 t} + c_{2} e^{2 t} + \frac{1}{3}

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

a, 5 d (a) = 6 d (a) + 9 d^{2}

y^{^{''}} + 4 y^{^{'}} + 4 = e^{- 2 x} ln (x)

\frac{3}{16} y^{^{''}} + y = 4 cos (7 t) + 6

\frac{d ^{2} y}{d x ^{2}} - 7 \frac{d y}{d x} + 9 y = x e^{x}

What is the solution for y^{''}-11y^'+18y=6 ?
The solution for y^{''}-11y^'+18y=6 is y=c_{1}e^{9t}+c_{2}e^{2t}+1/3