What is the solution for y^{''}+6y^'+5y=0,y(0)=1,y^'(0)=3 ?

The solution for y^{''}+6y^'+5y=0,y(0)=1,y^'(0)=3 is y=2e^{-t}-e^{-5t}

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y^{''}+6y^'+5y=0,y(0)=1,y^'(0)=3

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 3

y = 2 e^{- t} - e^{- 5 t}

Solution steps

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

Solve linear ODE:

y = 2 e^{- t} - e^{- 5 t}

y = 2 e^{- t} - e^{- 5 t}

y^{^{''}} + 4 y = 0, y^{^{'}} (0) = 2

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + \frac{1}{4} y = 0

100 y^{^{''}} + 200 y^{^{'}} + 101 y = 0

y^{^{''''}} - 8 y = 0

\frac{d ^{2} x}{d t ^{2}} + 8 \frac{d x}{d t} + 4 x = 0

What is the solution for y^{''}+6y^'+5y=0,y(0)=1,y^'(0)=3 ?
The solution for y^{''}+6y^'+5y=0,y(0)=1,y^'(0)=3 is y=2e^{-t}-e^{-5t}