What is the solution for y^{''''}-y^{''}+4y=0 ?

The solution for y^{''''}-y^{''}+4y=0 is

Popular >

y^{''''}-y^{''}+4y=0

y^{^{''''}} - y^{^{''}} + 4 y = 0

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2})) + e^{- \frac{5 t}{2}} (c_{3} cos (\frac{3 t}{2}) + c_{4} sin (\frac{3 t}{2}))

Solution steps

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

Solve linear ODE:

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2})) + e^{- \frac{5 t}{2}} (c_{3} cos (\frac{3 t}{2}) + c_{4} sin (\frac{3 t}{2}))

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2})) + e^{- \frac{5 t}{2}} (c_{3} cos (\frac{3 t}{2}) + c_{4} sin (\frac{3 t}{2}))

x^{^{''}} - 4 x^{^{'}} + 4 x = 0

6 y^{^{''}} + 17 y^{^{'}} - 3 y = 0

y^{^{''}} - 15 y^{^{'}} + 56 y = 0

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

y^{^{''}} + 2 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

What is the solution for y^{''''}-y^{''}+4y=0 ?
The solution for y^{''''}-y^{''}+4y=0 is