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

The solution for y^{''''}+2y^{''}=0 is y=c_{1}+c_{2}t+c_{3}cos(sqrt(2)t)+c_{4}sin(sqrt(2)t)

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

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

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

Solution steps

y^{(4)} (t) + 2 y^{''} (t) = 0

Solve linear ODE:

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

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

\int \frac{x + 2}{x ^{2} + 4 x} d x

\int \frac{1}{1 + 3 x} d x

y = sin (tan (3 x))

\int x^{2} e^{- 14 x} d x

\int_{5}^{8} 1

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