What is the solution for y^{'''}+4y^'=0,y(0)=1,y^'(0)=1,y^{''}(0)=1 ?

The solution for y^{'''}+4y^'=0,y(0)=1,y^'(0)=1,y^{''}(0)=1 is y=1+1/4-1/4 cos(2t)+1/2 sin(2t)

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

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

y = 1 + \frac{1}{4} - \frac{1}{4} cos (2 t) + \frac{1}{2} sin (2 t)

Solution steps

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

Solve linear ODE:

y = 1 + \frac{1}{4} - \frac{1}{4} cos (2 t) + \frac{1}{2} sin (2 t)

y = 1 + \frac{1}{4} - \frac{1}{4} cos (2 t) + \frac{1}{2} sin (2 t)

144 y^{^{''}} - 24 y^{^{'}} + y = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = - \frac{8}{5}, y^{^{'}} (0) = - 5

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 2, y (0) = - 7

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

x^{^{''}} + 6 x^{^{'}} + 3 x = 0

What is the solution for y^{'''}+4y^'=0,y(0)=1,y^'(0)=1,y^{''}(0)=1 ?
The solution for y^{'''}+4y^'=0,y(0)=1,y^'(0)=1,y^{''}(0)=1 is y=1+1/4-1/4 cos(2t)+1/2 sin(2t)