What is the solution for y^{''}+4y^'-21y=0 ?

The solution for y^{''}+4y^'-21y=0 is y=c_{1}e^{3t}+c_{2}e^{-7t}

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y^{''}+4y^'-21y=0

y^{^{''}} + 4 y^{^{'}} - 21 y = 0

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

Solution steps

y^{''} (t) + 4 y^{'} (t) - 21 y = 0

Solve linear ODE:

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

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

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(\frac{1}{x ^{2}} - \frac{3}{x ^{4}}) (x + 7 x^{3})

(5 y - 2 x) (\frac{d y}{d x}) - 2 y = 0

What is the solution for y^{''}+4y^'-21y=0 ?
The solution for y^{''}+4y^'-21y=0 is y=c_{1}e^{3t}+c_{2}e^{-7t}