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

The solution for y^{''}-y^'-4y=0 is y=c_{1}e^{((1+sqrt(17))t)/2}+c_{2}e^{((1-sqrt(17))t)/2}

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

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

y = c_{1} e^{\frac{( 1 + 17 ) t}{2}} + c_{2} e^{\frac{( 1 - 17 ) t}{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( 1 + 17 ) t}{2}} + c_{2} e^{\frac{( 1 - 17 ) t}{2}}

y = c_{1} e^{\frac{( 1 + 17 ) t}{2}} + c_{2} e^{\frac{( 1 - 17 ) t}{2}}

\frac{d}{d x} (e^{(x)})

x \to \infty lim (\frac{1}{x} + x)

4 e^{t}

x \to 4 lim (\frac{2}{( x + 4 ) ^{3}})

3 e^{- 2 t} cos (t)

What is the solution for y^{''}-y^'-4y=0 ?
The solution for y^{''}-y^'-4y=0 is y=c_{1}e^{((1+sqrt(17))t)/2}+c_{2}e^{((1-sqrt(17))t)/2}