Question

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⟩

Answer 1

$\mathbf r'(t)=\langle\sin9t,\sin6t,9t\rangle$

$\mathbf r(t)=\displaystyle\int\mathbf r'(t)\,\mathrm dt$

$\mathbf r(t)=\left\langle\displaystyle\int\sin9t\,\mathrm dt,\int\sin6t\,\mathrm dt,\int9t\,\mathrm dt\right\rangle$

$\displaystyle\int\sin9t\,\mathrm dt=\frac19\cos9t+C_1$

$\displaystyle\int\sin6t\,\mathrm dt=\frac16\cos6t+C_2$

$\displaystyle\int9t\,\mathrm dt=\frac92t^2+C_3$

With the initial condition $\mathbf r(0)=\langle4,6,3\rangle$, we find

$\dfrac19\cos0+C_1=4\implies C_1=\dfrac{35}9$

$\dfrac16\cos0+C_2=6\implies C_2=\dfrac{35}6$

$\dfrac92\cdot0^2+C_3=3\implies C_3=3$

So the particular solution to the IVP is

$\mathbf r(t)=\left\langle\dfrac19\cos9t+\dfrac{35}9,\dfrac16\cos6t+\dfrac{35}6,\dfrac92t^2+3\right\rangle$