What is the solution for y^{''}-10y^'+9y=27 ?

The solution for y^{''}-10y^'+9y=27 is y=c_{1}e^{9t}+c_{2}e^t+3

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y^{''}-10y^'+9y=27

y^{^{''}} - 10 y^{^{'}} + 9 y = 27

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

Solution steps

y^{''} (t) - 10 y^{'} (t) + 9 y = 27

Solve linear ODE:

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

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

\frac{d}{d x} (x^{2} - 4 x + 4)

\frac{d}{d x} (\frac{3}{x ^{5}})

x \to 0 + lim (ln (5 sin (x)))

x \to 2 lim ((x - 2)^{4} (5 x^{2}))

x y^{^{'}} + y = 7 x y^{2}, y (1) = 5

What is the solution for y^{''}-10y^'+9y=27 ?
The solution for y^{''}-10y^'+9y=27 is y=c_{1}e^{9t}+c_{2}e^t+3