What is the solution for y^{''}-2y^'-3y=0,y(0)=5 ?

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

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y^{''}-2y^'-3y=0,y(0)=5

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 5

y = (5 - c_{2}) e^{3 t} + c_{2} e^{- t}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = (5 - c_{2}) e^{3 t} + c_{2} e^{- t}

y = (5 - c_{2}) e^{3 t} + c_{2} e^{- t}

x^{^{'''}} - x^{^{''}} = 0, x (0) = 1, x^{^{'}} (0) = 3, x^{^{''}} (0) = 2

y^{^{''}} - 10 y^{^{'}} + 34 y = 0, y (0) = 5, y^{^{'}} (0) = - 2

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

4 y^{^{''}} + 28 y = 0, y (0) = 3, y^{^{'}} (0) = 7

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

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