What is the solution for y^{'''}+36y^'=0,y^'(0)=6 ?

The solution for y^{'''}+36y^'=0,y^'(0)=6 is y=c_{1}+c_{2}cos(6t)+sin(6t)

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y^{'''}+36y^'=0,y^'(0)=6

y^{^{'''}} + 36 y^{^{'}} = 0, y^{^{'}} (0) = 6

y = c_{1} + c_{2} cos (6 t) + sin (6 t)

Solution steps

y^{'''} (t) + 36 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} cos (6 t) + sin (6 t)

y = c_{1} + c_{2} cos (6 t) + sin (6 t)

y^{^{'''}} - 5 y^{^{''}} - 2 y^{^{'}} + 24 y = 0

f^{^{''}} (x) + f (x) = 0, f (0) = 1

f^{^{''}} (x) + f (x) = 0, f (0) = 0

\frac{d ^{2} x}{d t ^{2}} - 5 \frac{d x}{d t} + 4 x = 0

\frac{d ^{2} y}{d t ^{2}} + 8 \frac{d y}{d t} + 4 y = 0

What is the solution for y^{'''}+36y^'=0,y^'(0)=6 ?
The solution for y^{'''}+36y^'=0,y^'(0)=6 is y=c_{1}+c_{2}cos(6t)+sin(6t)