What is the solution for 15y^{''}+7y^'-2y=0 ?

The solution for 15y^{''}+7y^'-2y=0 is y=c_{1}e^{t/5}+c_{2}e^{-(2t)/3}

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15y^{''}+7y^'-2y=0

15 y^{^{''}} + 7 y^{^{'}} - 2 y = 0

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{- \frac{2 t}{3}}

Solution steps

15 y^{''} (t) + 7 y^{'} (t) - 2 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{- \frac{2 t}{3}}

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{- \frac{2 t}{3}}

\int sin (π x) cos (π x) d x

\frac{d}{d x} (cosh (a x))

\int \frac{2 x ^{2} + x + 2}{( x ^{2} + 1 ) ^{2}} d x

What is the solution for 15y^{''}+7y^'-2y=0 ?
The solution for 15y^{''}+7y^'-2y=0 is y=c_{1}e^{t/5}+c_{2}e^{-(2t)/3}