Using Phasors, determine i(t) for the following integro-differential equation:

$$ 10\int_{}^{} i\;dt + \frac{di}{dt} + 6i = 5\cos(5t + 22^{\circ}) $$ By inspection we can see that the angular frequency is: $$ \omega = 5\frac{rad}{s} $$ ...and from the intro to sinusoid and phasor calculus page we know that the derivative of i(t) is equivalent to the following in phasor form: $$ \frac{di}{dt} \iff j\omega \mathbb{I} $$ ...and that the integral of i(t) is equivalent to the following in phasor form: $$ \int_{}^{} i \; dt = \frac{\mathbb{I}}{j\omega} $$ ...where: $$ \mathbb{I} = phasor \; form \; of \; the \; sinusoid \; i(t) $$ This allows us to rewrite the equation as: $$ \frac{10\mathbb{I}}{j5} + j5\mathbb{I} + 6\mathbb{I} = 5\cos(5t+22^{\circ}) \qquad(Expression \; 1)$$

Note that the right side of the equation is equivalent to the following: $$ 5\cos(5t+22^{\circ}) = 5\angle22^{\circ}$$ ...which allows us to rewrite expression #1 as:

$$ \frac{10\mathbb{I}}{j5} + j5\mathbb{I} + 6\mathbb{I} = 5\angle22^{\circ} $$

Goal: To isolate to phasor term on the left side of the equation and convert it to i(t) in sinusoid form

$$ \mathbb{I} \Big( \frac{2}{j} + j5 + 6 \Big) = 5\angle22^{\circ} \qquad (Expression \; 2) $$

Recall from our rules for complex numbers in the review of complex numbers page that: $$ 2 \Big( \frac{1}{j} \Big) = -2j $$ ..which means that we can rewrite expression #2 as:

$$ \mathbb{I}(6 - j2 + j5) = 5\angle22^{\circ} $$ $$ \mathbb{I}(6 + 3j) = 5\angle22^{\circ} \qquad (Expression \; 3)$$

For the 6+3j term: $$ r = \sqrt{6^2+3^2} $$ $$ r = 3\sqrt{5} $$ $$ \phi_{ref} = tan^{-1} \Big( \frac{3}{6} \Big) \;\;\;,lies \; 1^{st} \; quadrant $$ $$ \phi_{ref} = 26.57^{\circ} = \phi $$ Therefore: $$ 6+3j = 3\sqrt{5} \angle 26.57^{\circ} $$ ...which allows us to rewrite expression #3 as:

$$ \mathbb{I} = \frac{ 5\angle22^{\circ}}{3\sqrt{5} \angle 26.57^{\circ}} \qquad (Expression \; 4) $$

Recall the property of division for complex numbers from our review of complex numbers page $$ \frac{z_1}{z_2} = \frac{r_1}{r_2} \angle (\phi_1 - \phi_2) $$ ...and use it on expression #4:

$$ \mathbb{I} = \Big( \frac{5}{3\sqrt{5}} \Big) \angle (22^{\circ} - 26.57^{\circ}) $$ $$ \mathbb{I} = \Big( \frac{\sqrt{5}}{3} \Big) \angle (-4.57^{\circ}) $$ ...which is equivalent to the following when expressed in exponential form: $$ \mathbb{I} = \Big( \frac{\sqrt{5}}{3} \Big)e^{j(-4.57^{\circ})} \qquad(Expression \; 5) $$ We now need to multiply expression #5 by e^jwt and take the real part in order to obtain the sinusoid i(t). Recall that omega equals 5. $$ i(t) = R_e \Big\{ \Big( \frac{\sqrt{5}}{3} \Big)e^{j(-4.57^{\circ})} e^{j5t} \Big\} $$ $$ i(t) = R_e \Big\{ \Big( \frac{\sqrt{5}}{3} \Big) e^{j(5t-4.57^{\circ})} \Big\} \qquad(Expression \; 6) $$

Recall Euhler's Identity: $$ e^{\pm j\phi} = \cos\phi \pm j\sin\phi $$ ...and apply it to expression #6:

$$ i(t) = R_e \Big\{ \frac{\sqrt{5}}{3}\cos(5t-4.57^{\circ}) + j\frac{\sqrt{5}}{3} \sin(5t-4.57^{\circ}) \Big\} $$ ...which we exaluate to obtain the:

Final Answer:

$$ i(t) = 0.745\cos(5t-4.57^{\circ}) $$