What is the solution for 25y^{''}-4y=0 ?

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

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25y^{''}-4y=0

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

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

Solution steps

25 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}}

e^{3 k}

\frac{d}{d x} ((3 x^{4} + 4 x)^{2} (5 x + 8)^{4})

t \to 0 lim (\frac{e ^{6 t} - 1}{t})

y = - 4 e^{- 4 x}

\int 7 cos (x) d x

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