Find the solution r(t)

of the differential equation with the given initial condition:

r′(t)=⟨sin5t,sin8t,3t⟩,r(0)=⟨8,2,9⟩

1 answer

To find the solution r(t) of the differential equation, we need to integrate each component of r'(t) with respect to t.

For the first component, integrating sin(5t) with respect to t gives us -(1/5)cos(5t) + C1, where C1 is the constant of integration.
For the second component, integrating sin(8t) with respect to t gives us -(1/8)cos(8t) + C2, where C2 is the constant of integration.
For the third component, integrating 3t with respect to t gives us (3/2)t^2 + C3, where C3 is the constant of integration.

Therefore, the solution r(t) is given by:
r(t) = ⟨-(1/5)cos(5t) + C1, -(1/8)cos(8t) + C2, (3/2)t^2 + C3⟩.

To find the specific solution r(t) with the initial condition r(0) = ⟨8, 2, 9⟩, we substitute t = 0 into the solution and solve for the constants C1, C2, and C3.

r(0) = ⟨-(1/5)cos(5*0) + C1, -(1/8)cos(8*0) + C2, (3/2)*0^2 + C3⟩
⟨8, 2, 9⟩ = ⟨-(1/5) + C1, -1/8 + C2, C3⟩.

By comparing the corresponding components, we get the equations:
-(1/5) + C1 = 8
-1/8 + C2 = 2
C3 = 9

Solving these equations, we find:
C1 = 8 + 1/5 = 41/5
C2 = 2 + 1/8 = 17/8
C3 = 9

Therefore, the specific solution r(t) with the initial condition r(0) = ⟨8, 2, 9⟩ is:
r(t) = ⟨-(1/5)cos(5t) + 41/5, -(1/8)cos(8t) + 17/8, (3/2)t^2 + 9⟩.