What is the solution for 25y^{''}-25y^'-6y=0 ?

The solution for 25y^{''}-25y^'-6y=0 is y=c_{1}e^{(6t)/5}+c_{2}e^{-t/5}

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25y^{''}-25y^'-6y=0

25 y^{^{''}} - 25 y^{^{'}} - 6 y = 0

y = c_{1} e^{\frac{6 t}{5}} + c_{2} e^{- \frac{t}{5}}

Solution steps

25 y^{''} (t) - 25 y^{'} (t) - 6 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{6 t}{5}} + c_{2} e^{- \frac{t}{5}}

y = c_{1} e^{\frac{6 t}{5}} + c_{2} e^{- \frac{t}{5}}

\frac{d}{d x} (\frac{e ^{x} ( x ^{2} - 4 x + 5 )}{( 1 - x ) ^{3}})

\int tan (x) d x

\int \frac{1}{x 25 - x ^{2}} d x

\int 5 x^{3} d x

\int 2 sec^{2} (x) tan^{3} (x) d x

What is the solution for 25y^{''}-25y^'-6y=0 ?
The solution for 25y^{''}-25y^'-6y=0 is y=c_{1}e^{(6t)/5}+c_{2}e^{-t/5}