What is the solution for 4y^{''}-4y^'-3y=0,y(0)=5 ?

The solution for 4y^{''}-4y^'-3y=0,y(0)=5 is y=(5-c_{2})e^{(3t)/2}+c_{2}e^{-t/2}

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4y^{''}-4y^'-3y=0,y(0)=5

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 5

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = 8, y^{^{'}} (0) = - 14

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

y^{^{''}} - 25 y = 0, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} + 6 y^{^{'}} + 34 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

9 y^{^{''}} + 16 y = 0

What is the solution for 4y^{''}-4y^'-3y=0,y(0)=5 ?
The solution for 4y^{''}-4y^'-3y=0,y(0)=5 is y=(5-c_{2})e^{(3t)/2}+c_{2}e^{-t/2}