Question

(EDO) Alguém saberia calcular a seguinte equação diferencial: $\frac{dy}{dt} =e^{t} cost$?

Answer 1

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$\displaystyle\sf\int e^t~cos~t~dt\\\sf u=cos~t\implies du=-sen~t~dt\\\sf dv=e^t\implies v=e^t\\\displaystyle\sf\int e^t~cos~t~dt=e^t cos~t-\int e^t\cdot(-sen~t)~dt\\\displaystyle\sf\int e^t~cos~t~dt=e^t~cos~t+\int e^t~sen~t~dt\\\sf u_1=sen~t\implies du_1= cos~t~dt\\\sf dv_1=e^t\implies v_1=e^t\\\boxed{\displaystyle\sf\int e^t~sen~t~dt=e^t sen~t-\int e^t~cos~t~dt}$

$\displaystyle\sf\int e^t~cos~t~dt=e^t cos~t+e^t~sen~t-\int e^t~cos~t~dt\\\displaystyle\sf\int e^t~cos~t~dt+\int e^t~cos~t~dt=e^tcos~t+e^t~sen~t\\\displaystyle\sf2\int e^t~cos~t~dt=e^t(cos~t+sen~t)\\\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int e^t~cos~t~dt=\dfrac{1}{2}e^t(cos~t+sen~t)+k}}}}$

$\sf \dfrac{dy}{dt}=e^tcos~t\\\sf dy=e^tcos~t~dt\\\displaystyle\sf\int dy=\int e^t~cos~t~dt\\\sf y(t)=\dfrac{1}{2}e^t(cos~t+sen~t)+k$