What is the solution for y^{''}+3y^'-40y=0 ?

The solution for y^{''}+3y^'-40y=0 is y=c_{1}e^{5t}+c_{2}e^{-8t}

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y^{''}+3y^'-40y=0

y^{^{''}} + 3 y^{^{'}} - 40 y = 0

y = c_{1} e^{5 t} + c_{2} e^{- 8 t}

Solution steps

y^{''} (t) + 3 y^{'} (t) - 40 y = 0

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- 8 t}

y = c_{1} e^{5 t} + c_{2} e^{- 8 t}

\int \frac{- 2}{t + 1} d t

\frac{1200}{s ^{2} ( s + 10 ) ^{2}}

\frac{2 x ^{3} - 6}{x ^{4}}

\frac{d}{d x} (\frac{x cos ( x ) - sin ( x )}{x ^{2}})

\frac{\partial}{\partial x} (30 x^{5} y^{4} - 20 x^{4} y^{3})

What is the solution for y^{''}+3y^'-40y=0 ?
The solution for y^{''}+3y^'-40y=0 is y=c_{1}e^{5t}+c_{2}e^{-8t}