What is the solution for 9y^{''}+9y^'-4y=0 ?

The solution for 9y^{''}+9y^'-4y=0 is y=c_{1}e^{t/3}+c_{2}e^{-(4t)/3}

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9y^{''}+9y^'-4y=0

9 y^{^{''}} + 9 y^{^{'}} - 4 y = 0

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

Solution steps

9 y^{''} (t) + 9 y^{'} (t) - 4 y = 0

Solve linear ODE:

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

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

x \to 0 lim (x cos (\frac{9}{x}))

\frac{d ^{2}}{d x ^{2}} (\frac{7}{8 t ^{5}})

x \to 1 - lim (x - 2)

y = 1 + x, y = \frac{( 3 + x )}{3}

\int 6 - x d x

What is the solution for 9y^{''}+9y^'-4y=0 ?
The solution for 9y^{''}+9y^'-4y=0 is y=c_{1}e^{t/3}+c_{2}e^{-(4t)/3}