What is the solution for y^{''}-6y^'+9y=18 ?

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

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y^{''}-6y^'+9y=18

y^{^{''}} - 6 y^{^{'}} + 9 y = 18

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

Solution steps

y^{''} (t) - 6 y^{'} (t) + 9 y = 18

Solve linear ODE:

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

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

y^{^{''}} - y = 8 t e^{t} + 2 e^{t}

y^{^{''}} + 18 y = 3 e^{8 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{e ^{2 x}}{1 + x}

q^{^{''}} + 3 q^{^{'}} + 2 q = e^{- t} + e^{- 2 t}

y^{^{''}} - 14 y^{^{'}} + 40 y = 26 + 40 x

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