What is the solution for y^{''}-2sqrt(2)y^'+2y=0 ?

The solution for y^{''}-2sqrt(2)y^'+2y=0 is y=c_{1}e^{sqrt(2)t}+c_{2}te^{sqrt(2)t}

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y^{''}-2sqrt(2)y^'+2y=0

y^{^{''}} - 22 y^{^{'}} + 2 y = 0

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

Solution steps

y^{''} (t) - 22 y^{'} (t) + 2 y = 0

Solve linear ODE:

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

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

1 y^{^{''}} + 10 y^{^{'}} + \frac{32}{10} y = 0

y^{^{''}} - 20 y^{^{'}} - a y = 0, y (0) = 1, y^{^{'}} (0) = 15

y^{^{''}} - y^{^{'}} + \frac{y}{4} = 0

y^{^{'''}} + y^{^{''}} + 2 y^{^{'}} + 2 y = 0

\frac{d ^{2} f}{d x ^{2}} + 4 f (x) = 0

What is the solution for y^{''}-2sqrt(2)y^'+2y=0 ?
The solution for y^{''}-2sqrt(2)y^'+2y=0 is y=c_{1}e^{sqrt(2)t}+c_{2}te^{sqrt(2)t}