What is the solution for y^{''}-4y^'-4y=0 ?

The solution for y^{''}-4y^'-4y=0 is y=c_{1}e^{2(1+sqrt(2))t}+c_{2}e^{2(1-sqrt(2))t}

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y^{''}-4y^'-4y=0

y^{^{''}} - 4 y^{^{'}} - 4 y = 0

y = c_{1} e^{2 (1 + 2) t} + c_{2} e^{2 (1 - 2) t}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 4 y = 0

Solve linear ODE:

y = c_{1} e^{2 (1 + 2) t} + c_{2} e^{2 (1 - 2) t}

y = c_{1} e^{2 (1 + 2) t} + c_{2} e^{2 (1 - 2) t}

\int_{0}^{π} x^{2} sin^{2} (x) d x

\frac{sin ( x ^{2} - 5 )}{2 x}

x \to 1 lim (\frac{x ^{2} - 1}{x + 1})

\int u cos (2 u) d u

\frac{df}{d x} f (x) = cos (2 x)

What is the solution for y^{''}-4y^'-4y=0 ?
The solution for y^{''}-4y^'-4y=0 is y=c_{1}e^{2(1+sqrt(2))t}+c_{2}e^{2(1-sqrt(2))t}