What is the solution for 4y^{''}+7y^'+3y=0 ?

The solution for 4y^{''}+7y^'+3y=0 is y=c_{1}e^{-(3t)/4}+c_{2}e^{-t}

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4y^{''}+7y^'+3y=0

4 y^{^{''}} + 7 y^{^{'}} + 3 y = 0

y = c_{1} e^{- \frac{3 t}{4}} + c_{2} e^{- t}

Solution steps

4 y^{''} (t) + 7 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{3 t}{4}} + c_{2} e^{- t}

y = c_{1} e^{- \frac{3 t}{4}} + c_{2} e^{- t}

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (- 1) = 5, y^{^{'}} (- 1) = 2

c y^{^{''}} + ab y^{^{'}} = 0, y (0) = 1, y (1.2) = 0

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

y^{^{''}} + 2 y^{^{'}} + 65 y = 0, y (0) = - 1, y^{^{'}} (0) = 33

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 5

What is the solution for 4y^{''}+7y^'+3y=0 ?
The solution for 4y^{''}+7y^'+3y=0 is y=c_{1}e^{-(3t)/4}+c_{2}e^{-t}