What is the solution for 5y^{''}+3y^'-y=0 ?

The solution for 5y^{''}+3y^'-y=0 is y=c_{1}e^{((-3+sqrt(29))t)/(10)}+c_{2}e^{((-3-sqrt(29))t)/(10)}

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5y^{''}+3y^'-y=0

5 y^{^{''}} + 3 y^{^{'}} - y = 0

y = c_{1} e^{\frac{( - 3 + 29 ) t}{10}} + c_{2} e^{\frac{( - 3 - 29 ) t}{10}}

Solution steps

5 y^{''} (t) + 3 y^{'} (t) - y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( - 3 + 29 ) t}{10}} + c_{2} e^{\frac{( - 3 - 29 ) t}{10}}

y = c_{1} e^{\frac{( - 3 + 29 ) t}{10}} + c_{2} e^{\frac{( - 3 - 29 ) t}{10}}

\frac{d ^{2} y}{d t ^{2}} - 1 \frac{d y}{d t} = 0

x^{^{''}} = - 4 x

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.45

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.46

y^{^{''}} - 6 y^{^{'}} + y = 0

What is the solution for 5y^{''}+3y^'-y=0 ?
The solution for 5y^{''}+3y^'-y=0 is y=c_{1}e^{((-3+sqrt(29))t)/(10)}+c_{2}e^{((-3-sqrt(29))t)/(10)}