What is the solution for 25y^{''}-16y=0 ?

The solution for 25y^{''}-16y=0 is y=c_{1}e^{(4t)/5}+c_{2}e^{-(4t)/5}

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25y^{''}-16y=0

25 y^{^{''}} - 16 y = 0

y = c_{1} e^{\frac{4 t}{5}} + c_{2} e^{- \frac{4 t}{5}}

Solution steps

25 y^{''} (t) - 16 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{4 t}{5}} + c_{2} e^{- \frac{4 t}{5}}

y = c_{1} e^{\frac{4 t}{5}} + c_{2} e^{- \frac{4 t}{5}}

y^{^{''}} + 6 y^{^{'}} + 18 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 9

x^{^{''}} - 8 x^{^{'}} + 16 x = 0

\frac{d ^{3} x}{d t ^{3}} - \frac{d ^{2} x}{d t ^{2}} - 4 x = 0

y^{^{''}} + 3 y^{^{'}} + \frac{9}{4} y = 0

y^{^{''}} - y^{^{'}} - 12 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

What is the solution for 25y^{''}-16y=0 ?
The solution for 25y^{''}-16y=0 is y=c_{1}e^{(4t)/5}+c_{2}e^{-(4t)/5}