What is the solution for y^{''}+5y^'-y=0 ?

The solution for y^{''}+5y^'-y=0 is y=c_{1}e^{((-5+sqrt(29))t)/2}+c_{2}e^{((-5-sqrt(29))t)/2}

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y^{''}+5y^'-y=0

y^{^{''}} + 5 y^{^{'}} - y = 0

y = c_{1} e^{\frac{( - 5 + 29 ) t}{2}} + c_{2} e^{\frac{( - 5 - 29 ) t}{2}}

Solution steps

y^{''} (t) + 5 y^{'} (t) - y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 5 + 29 ) t}{2}} + c_{2} e^{\frac{( - 5 - 29 ) t}{2}}

y = c_{1} e^{\frac{( - 5 + 29 ) t}{2}} + c_{2} e^{\frac{( - 5 - 29 ) t}{2}}

y^{^{'''}} + 9 y^{^{''}} + 20 y^{^{'}} = 0

y^{^{''}} + 8 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{'''}} - 2 y^{^{'}} - 13 y^{^{'}} - 10 y = 0

\frac{d ^{2} y}{d x ^{2}} + 0.6 \frac{d y}{d x} + 8 y = 0

y^{^{''}} + 7 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 1

What is the solution for y^{''}+5y^'-y=0 ?
The solution for y^{''}+5y^'-y=0 is y=c_{1}e^{((-5+sqrt(29))t)/2}+c_{2}e^{((-5-sqrt(29))t)/2}