What is the solution for 15y^{''}-7y^'-4y=0 ?

The solution for 15y^{''}-7y^'-4y=0 is y=c_{1}e^{(4t)/5}+c_{2}e^{-t/3}

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15y^{''}-7y^'-4y=0

15 y^{^{''}} - 7 y^{^{'}} - 4 y = 0

y = c_{1} e^{\frac{4 t}{5}} + c_{2} e^{- \frac{t}{3}}

Solution steps

15 y^{''} (t) - 7 y^{'} (t) - 4 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{4 t}{5}} + c_{2} e^{- \frac{t}{3}}

y = c_{1} e^{\frac{4 t}{5}} + c_{2} e^{- \frac{t}{3}}

t \to 0 lim (tan (4 t))

\int 2 t - 3 sin (t) d t

\frac{d}{d x} (6 x^{2} - 7 x)

\frac{d y}{d x} = \frac{3 y}{x}

\int x^{- 4} (x + 1) d x

What is the solution for 15y^{''}-7y^'-4y=0 ?
The solution for 15y^{''}-7y^'-4y=0 is y=c_{1}e^{(4t)/5}+c_{2}e^{-t/3}