I am learning Signals and Systems by watching videos on-line and using the Alan V Oppenheim book. The current video I am watching has a portion missing out of it, the professor starts a problem of but the working out is missing. On the next video the solution is presented however I am unable to fill in the missing blanks . The video is here

I will write the question out and show my working out. If some one could show me where I have gone I will really appreciate it.

The question is to solve the following FODE.

y' + 2y= Kcos wt (u(t)) where y is the first derivative.

Solve using particular integral + the complementary solution.

The input can be expressed in the exponential form: Re[K* e^jwt]

PI= Assume solution for y that is similar to the input: Re[Y*e^jwt], substituting this in for y is

jwYe^jwt + 2Ye^jwt=Ke^jwt dividing by e^jwt.

jwY+2Y=K which makes Y = K/(jw+2) the absolute magnitude of |y| is: k/√(2+w^2) and ∅= -tan w/2.

A complex number can be written as |A| e^j∅ here A is |Y|e^jwt. is that is done the result is:

|Y|e^jwte^j∅ = |Y|e^j(wt+∅) [ in the video the prof has written Ye^j(wt-∅) which doesn't matter in this case as we know the input is real and even.

Returning the cosine, the PI = |Y| cos wt+cos∅.

The complementary function to the input assuming the input is 0 and y= Ae^mt.

substituting this in and working out m as -2 we have y= Ae^-2t.

I kind of got stuck after this, the solution is:

y(t)= Ye^-2t + k/(√4+w^2)[ cos(wt-∅) - cos ∅e-2t] u(t).

Where did the cos ∅e^2t come from?

I a have not been on this forum for a long time and dont know where the latex code option is so I apologise in advance for not using it.