What is the solution for y^{''}+3y^'-18y=0 ?

The solution for y^{''}+3y^'-18y=0 is y=c_{1}e^{3t}+c_{2}e^{-6t}

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y^{''}+3y^'-18y=0

y^{^{''}} + 3 y^{^{'}} - 18 y = 0

y = c_{1} e^{3 t} + c_{2} e^{- 6 t}

Solution steps

y^{''} (t) + 3 y^{'} (t) - 18 y = 0

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- 6 t}

y = c_{1} e^{3 t} + c_{2} e^{- 6 t}

\frac{d}{d x} (\frac{2 sin ( x )}{3 + cos ( x )})

y = 96 x^{2} + 72 x + 12

y^{^{'}} cos (x) = y sin (x) + sin (103 x)

(x + 2 x) (2 x + 1)

\frac{d y}{d x}, 1 + ln (9 x y) = e^{(9 x - y)}

What is the solution for y^{''}+3y^'-18y=0 ?
The solution for y^{''}+3y^'-18y=0 is y=c_{1}e^{3t}+c_{2}e^{-6t}