What is the solution for 32y^{''}+20y^'-3y=0 ?

The solution for 32y^{''}+20y^'-3y=0 is y=c_{1}e^{t/8}+c_{2}e^{-(3t)/4}

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32y^{''}+20y^'-3y=0

32 y^{^{''}} + 20 y^{^{'}} - 3 y = 0

y = c_{1} e^{\frac{t}{8}} + c_{2} e^{- \frac{3 t}{4}}

Solution steps

32 y^{''} (t) + 20 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{8}} + c_{2} e^{- \frac{3 t}{4}}

y = c_{1} e^{\frac{t}{8}} + c_{2} e^{- \frac{3 t}{4}}

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What is the solution for 32y^{''}+20y^'-3y=0 ?
The solution for 32y^{''}+20y^'-3y=0 is y=c_{1}e^{t/8}+c_{2}e^{-(3t)/4}