What is the solution for y^{''}+6y^'+5y=0,y(0)=0,y^'(0)=3 ?

The solution for y^{''}+6y^'+5y=0,y(0)=0,y^'(0)=3 is y= 3/4 e^{-t}-3/4 e^{-5t}

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y^{''}+6y^'+5y=0,y(0)=0,y^'(0)=3

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y = \frac{3}{4} e^{- t} - \frac{3}{4} e^{- 5 t}

Solution steps

y^{''} (t) + 6 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = \frac{3}{4} e^{- t} - \frac{3}{4} e^{- 5 t}

y = \frac{3}{4} e^{- t} - \frac{3}{4} e^{- 5 t}

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y^{^{''}} - 4 y^{^{'}} + 13 y = 2 e^{x}

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What is the solution for y^{''}+6y^'+5y=0,y(0)=0,y^'(0)=3 ?
The solution for y^{''}+6y^'+5y=0,y(0)=0,y^'(0)=3 is y= 3/4 e^{-t}-3/4 e^{-5t}