What is the solution for 5y^{''}-4y=0 ?

The solution for 5y^{''}-4y=0 is y=c_{1}e^{(2t)/(sqrt(5))}+c_{2}e^{-(2t)/(sqrt(5))}

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5y^{''}-4y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

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What is the solution for 5y^{''}-4y=0 ?
The solution for 5y^{''}-4y=0 is y=c_{1}e^{(2t)/(sqrt(5))}+c_{2}e^{-(2t)/(sqrt(5))}