What is the solution for 2y^{''}=3y ?

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

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2y^{''}=3y

2 y^{^{''}} = 3 y

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

Solution steps

2 y^{''} (t) = 3 y

Solve linear ODE:

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

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

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

36 y^{^{''}} + 156 y^{^{'}} + 169 y = 0, y (0) = 9, y^{^{'}} (0) = 4

y^{^{''}} - 6 y = 0, y (0) = 0, y (1) = 0

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

\frac{d ^{2} y}{d x ^{2}} + 10 \frac{d y}{d x} + 61 y = 0

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