What is the solution for (d^2u)/(dy^2)-4u+6=0 ?

The solution for (d^2u)/(dy^2)-4u+6=0 is u=c_{1}e^{2y}+c_{2}e^{-2y}+3/2

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(d^2u)/(dy^2)-4u+6=0

\frac{d ^{2} u}{d y ^{2}} - 4 u + 6 = 0

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

Solution steps

\frac{d ^{2} u}{d y ^{2}} - 4 u + 6 = 0

Solve linear ODE:

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

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

y^{^{''}} - 16 y = 32 + 45 e^{x}, y (0) = 0

y^{^{'''}} - 81 y^{^{'}} = t e^{- 9 t} + 10 cos (9 t)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y - e^{x} = 0

y^{^{''}} + 16 y = 5 sec (4 t)

y^{^{''}} - 3 y^{^{'}} - 4 y = e^{4 x}

What is the solution for (d^2u)/(dy^2)-4u+6=0 ?
The solution for (d^2u)/(dy^2)-4u+6=0 is u=c_{1}e^{2y}+c_{2}e^{-2y}+3/2