What is the solution for (D^4-D^2)y=0 ?

The solution for (D^4-D^2)y=0 is y=c_{1}e^x+c_{2}e^{-x}+c_{3}+c_{4}x

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(D^4-D^2)y=0

(D^{4} - D^{2}) y = 0

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} + c_{4} x

Solution steps

(D^{4} - D^{2}) y = 0

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} + c_{4} x

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} + c_{4} x

\frac{1}{2} x^{^{''}} + 3 x^{^{'}} + 4 x = 0, x (0) = 3, x^{^{'}} (0) = 0

y^{^{'''}} - 7 y^{^{''}} + 10 y = 0, y (0) = 7

y^{^{''}} + y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - y^{^{'}} = 0, y (0) = cos (x)

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

What is the solution for (D^4-D^2)y=0 ?
The solution for (D^4-D^2)y=0 is y=c_{1}e^x+c_{2}e^{-x}+c_{3}+c_{4}x