What is the solution for y^{''}+4y^'+3y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+4y^'+3y=0,y(0)=1,y^'(0)=0 is y=(1+1/2)e^{-t}-1/2 e^{-3t}

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y^{''}+4y^'+3y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 4 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = (1 + \frac{1}{2}) e^{- t} - \frac{1}{2} e^{- 3 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = (1 + \frac{1}{2}) e^{- t} - \frac{1}{2} e^{- 3 t}

y = (1 + \frac{1}{2}) e^{- t} - \frac{1}{2} e^{- 3 t}

2 y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

x^{^{''}} + x^{^{'}} + 2 x = 0

y^{^{''}} + y^{^{'}} - 20 y = 0, y (0) = 1, y (0) = 22

20 y^{^{''}} + 60 y = 0

\frac{d ^{2} y}{d x ^{2}} - 36 y = 0

What is the solution for y^{''}+4y^'+3y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+4y^'+3y=0,y(0)=1,y^'(0)=0 is y=(1+1/2)e^{-t}-1/2 e^{-3t}