Find y as a function of x if y'''+64y'=0, y(0)=1 y'(0)=24 y''(0)=64 y(x)= ?

Answer 1

The given equation is

y''' + 64y' = 0

Which is a 3rd order homogeneous ODE. The characteristic equation corresponding to this ODE is

r³ + 64r = 0

r = 0, 8i, -8i

So, the solution is

y = C₁ + C₂e^i8x + C₃e^-i8x

or

y = C₁ + C₂cos(8x) + C₃sin(8x)

which implies

y' = - 8C₂sin(8x) + 8C₃cos(8x)

y'' = - 64C₂cos(8x) - 64C₃sin(8x)

Using the condition y''(0) = 64, we get C₂ = -1

Using the condition y'(0) = 24, we get C₃ = 3

Finally, using the condition y(0) = 1, we get C₁ = 2

Therefore, the solution is

y = 2 - cos(8x) + 3sin(8x)