Find the solution r(t)r(t) of the differential equation with the given initial condition: r′(t)=⟨sin9t, sin6t,9t⟩,r(0)=⟨4,6,3⟩

Here we want to solve a differential equation:

r'(t) = < sin(9t), sin(6t), 9t >

With the restriction: r(0)=⟨4,6,3⟩

Remember that:

r'(t) = dr(t)/dt

Also, remember the general rules:

and:

Then using these rules we can get the antiderivatives of r(x) to get:

r(t) = < -cos(9t)/9 + c₁, -cos(6t)/6 + c₂, 9t²/2 + c₃ >

Where c₁, c₂, and c₃ are the constants of integration.

To find these values, we can use the given restriction:

r(0) = <4, 6, 3> = <  -cos(0)/9 + c₁, -cos(0)/6 + c₂, 0 + c₃ >

                        = <  - 1/9 + c₁, -1/6 + c₂,  c₃ >

Then we have 3 equations:

first component:

-1/9 + c₁ = 4

then

c₁ = 4 + 1/9 = 37/9

Second component:

- 1/6 + c₂ = 6

then

c₂ = 6 + 1/6 = 37/6

Third component:

c₃ = 3

Now we can conclude that the function is:

r(t) = < -cos(9t)/9 + 37/9, -cos(6t)/6 + 37/6 , 9t²/2 + 3 >

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With the initial condition , we find

So the particular solution to the IVP is

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