Question

Find the solution r(t)r(t) of the differential equation with the given initial condition: r′(t)=⟨sin4t,sin7t,9t⟩,r(0)=⟨9,8,3⟩

Answer 1

$\mathbf r'(t)=\langle\sin4t,\sin7t,9t\rangle$
$\implies\mathbf r(t)=\displaystyle\int\mathbf r'(t)\,\mathrm dt$
$\mathbf r(t)=\displaystyle\left\langle\int\sin4t\,\mathrm dt,\int\sin7t\,\mathrm dt,\int9t\,\mathrm dt\right\rangle$
$\mathbf r(t)=\left\langle-\dfrac14\cos4t+C_1,-\dfrac17\cos7t+C_2,-\dfrac92t^2+C_3\right\rangle$

With $\mathbf r(0)=\langle9,8,3\rangle$, we have

$\langle9,8,3\rangle=\left\langle-\dfrac14+C_1,-\dfrac17+C_2,C_3\right\rangle$
$\implies C_1=\dfrac{37}4,C_2=\dfrac{57}7,C_3=8$
$\implies\mathbf r(t)=\left\langle-\dfrac14\cos4t+\dfrac{37}4,-\dfrac17\cos7t+\dfrac{57}7,-\dfrac92t^2+8\right\rangle$