What is the solution for 24y^{''}+13y^'-7y=0 ?

The solution for 24y^{''}+13y^'-7y=0 is y=c_{1}e^{t/3}+c_{2}e^{-(7t)/8}

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24y^{''}+13y^'-7y=0

24 y^{^{''}} + 13 y^{^{'}} - 7 y = 0

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{- \frac{7 t}{8}}

Solution steps

24 y^{''} (t) + 13 y^{'} (t) - 7 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{- \frac{7 t}{8}}

y = c_{1} e^{\frac{t}{3}} + c_{2} e^{- \frac{7 t}{8}}

\int \frac{1}{( 1 + y ^{2} ) ^{\frac{3}{2}}} d y

f (x) = x ln (x + 1)

\frac{d}{d x} (sin (x) cos (x))

\frac{\partial}{\partial x} (\frac{7 x y}{x ^{2} + y ^{4}})

x \to 1 lim ((x - 1)^{(x - 1)})

What is the solution for 24y^{''}+13y^'-7y=0 ?
The solution for 24y^{''}+13y^'-7y=0 is y=c_{1}e^{t/3}+c_{2}e^{-(7t)/8}