What is the solution for y^{''}+4y^'+3y=0,y(0)=2,y^'(0)=-1 ?

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

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y^{''}+4y^'+3y=0,y(0)=2,y^'(0)=-1

y^{^{''}} + 4 y^{^{'}} + 3 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

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

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

2 y d y = \frac{5}{x} d x

\frac{d}{d x} (\frac{x + 1}{x + 3})

\frac{\partial}{\partial x} (\frac{2}{3} y^{- \frac{1}{3}})

y^{^{'}} = (x + y - 1)^{2}, y (0) = 0

\frac{d}{d x} (\frac{4 ( - x ^{2} - 4 )}{( x ^{2} - 4 ) ^{2}})

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