What is the solution for y^{''}+3/2 y^'+3/8 y=0 ?

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

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y^{''}+3/2 y^'+3/8 y=0

y^{^{''}} + \frac{3}{2} y^{^{'}} + \frac{3}{8} y = 0

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

Solution steps

y^{''} (t) + \frac{3}{2} y^{'} (t) + \frac{3}{8} y = 0

Solve linear ODE:

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

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

5 \frac{d ^{2} y}{d x ^{2}} - 7 \frac{d y}{d x} - 90 y = 0

y^{^{''}} - 9 y = 0, y^{^{'}} (0) = 0

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

\frac{d ^{2} y}{d x ^{2}} - 7 \frac{d y}{d x} = 0

(\frac{1}{9} d^{2} - 8 d - 2) y = 0

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