What is the solution for y^{'''}+9y^'=0,y^'(0)=-21 ?

The solution for y^{'''}+9y^'=0,y^'(0)=-21 is y=c_{1}+c_{2}cos(3t)-7sin(3t)

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y^{'''}+9y^'=0,y^'(0)=-21

y^{^{'''}} + 9 y^{^{'}} = 0, y^{^{'}} (0) = - 21

y = c_{1} + c_{2} cos (3 t) - 7 sin (3 t)

Solution steps

y^{'''} (t) + 9 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} cos (3 t) - 7 sin (3 t)

y = c_{1} + c_{2} cos (3 t) - 7 sin (3 t)

16 y^{^{''}} + 8 y^{^{'}} + 145 y = 0

2 \frac{d ^{2} y}{d x ^{2}} + 20 \frac{d y}{d x} + 51 y = 0

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

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

9 y^{^{''}} - 12 y^{^{'}} + 40 y = 0, y (0) = 8, y^{^{'}} (0) = 3

What is the solution for y^{'''}+9y^'=0,y^'(0)=-21 ?
The solution for y^{'''}+9y^'=0,y^'(0)=-21 is y=c_{1}+c_{2}cos(3t)-7sin(3t)