What is the solution for 12y^{''}-5y^'-3y=0,y(0)=2 ?

The solution for 12y^{''}-5y^'-3y=0,y(0)=2 is y=(2-c_{2})e^{(3t)/4}+c_{2}e^{-t/3}

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12y^{''}-5y^'-3y=0,y(0)=2

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

y = (2 - c_{2}) e^{\frac{3 t}{4}} + c_{2} e^{- \frac{t}{3}}

Solution steps

12 y^{''} (t) - 5 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = (2 - c_{2}) e^{\frac{3 t}{4}} + c_{2} e^{- \frac{t}{3}}

y = (2 - c_{2}) e^{\frac{3 t}{4}} + c_{2} e^{- \frac{t}{3}}

y^{^{''}} + 64 y = 0, y (0) = 4, y^{^{'}} (0) = 7

y^{^{''}} + b y^{^{'}} + 8 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{'}} + 4 y^{^{''}} = 0

y^{^{''''}} + 2 y^{^{'''}} + 4 y^{^{''}} + 8 y^{^{'}} = 0

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

What is the solution for 12y^{''}-5y^'-3y=0,y(0)=2 ?
The solution for 12y^{''}-5y^'-3y=0,y(0)=2 is y=(2-c_{2})e^{(3t)/4}+c_{2}e^{-t/3}