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

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

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

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

y = c_{1} + c_{2} x + c_{3} cos (x) + c_{4} sin (x)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} x + c_{3} cos (x) + c_{4} sin (x)

y = c_{1} + c_{2} x + c_{3} cos (x) + c_{4} sin (x)

\frac{d ^{4} y}{d t ^{4}} - y^{^{'''}} - y^{^{''}} + y^{^{'}} = 0

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 0

2 u^{^{''}} + u^{^{'}} + 3 H (t) = 0

y^{^{''}} + 2 y^{^{'}} - 10 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

2 y^{^{''}} - 0 y^{^{'}} + 18 y = 0

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