What is the solution for 25y^{''}+15y^'-4y=0 ?

The solution for 25y^{''}+15y^'-4y=0 is y=c_{1}e^{t/5}+c_{2}e^{-(4t)/5}

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25y^{''}+15y^'-4y=0

25 y^{^{''}} + 15 y^{^{'}} - 4 y = 0

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

Solution steps

25 y^{''} (t) + 15 y^{'} (t) - 4 y = 0

Solve linear ODE:

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

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

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What is the solution for 25y^{''}+15y^'-4y=0 ?
The solution for 25y^{''}+15y^'-4y=0 is y=c_{1}e^{t/5}+c_{2}e^{-(4t)/5}