What is the solution for y^{''}+4y^'-32y=0 ?

The solution for y^{''}+4y^'-32y=0 is y=c_{1}e^{4t}+c_{2}e^{-8t}

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y^{''}+4y^'-32y=0

y^{^{''}} + 4 y^{^{'}} - 32 y = 0

y = c_{1} e^{4 t} + c_{2} e^{- 8 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) - 32 y = 0

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- 8 t}

y = c_{1} e^{4 t} + c_{2} e^{- 8 t}

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\frac{( s + 1 )}{s ^{2} + s + 1}

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\int \frac{x ^{3} + 7}{x ^{2} + 5 x + 6} d x

What is the solution for y^{''}+4y^'-32y=0 ?
The solution for y^{''}+4y^'-32y=0 is y=c_{1}e^{4t}+c_{2}e^{-8t}