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⟩

Respuesta :

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

With [tex]\mathbf r(0)=\langle9,8,3\rangle[/tex], we have

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