What is the solution for 40y^{''}+7y^'-3y=0 ?

The solution for 40y^{''}+7y^'-3y=0 is y=c_{1}e^{t/5}+c_{2}e^{-(3t)/8}

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40y^{''}+7y^'-3y=0

40 y^{^{''}} + 7 y^{^{'}} - 3 y = 0

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{- \frac{3 t}{8}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{- \frac{3 t}{8}}

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{- \frac{3 t}{8}}

8 y^{^{''}} + 8 y^{^{'}} + \frac{1}{4} y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + k^{2} \cdot y = 0

y^{^{''}} = y

y^{^{''}} - 7 y^{^{'}} - 8 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''''}} - 8 y^{^{'''}} + 18 y^{^{''}} - 27 y = 0

What is the solution for 40y^{''}+7y^'-3y=0 ?
The solution for 40y^{''}+7y^'-3y=0 is y=c_{1}e^{t/5}+c_{2}e^{-(3t)/8}