particular solution to y'+2y=cos(10t) From "Circuit Analysis Demystified", David McMahon, 2008, Chapter 6, page 131, Quiz question 5.
$$y' + 2y = \cos (10t)$$
Following the method from Example 6-11 in the book.
I guess that the solution has the form:
$$y = Acos(10t + \phi)$$
$$y' = -10Asin(10t + \phi)$$
substituting this into the DE:
$$-10Asin(10t + \phi) + 2Acos(10t + \phi) = \cos(10t)$$
Now I use the following trig identities to factor the DE:
$$\begin{aligned} \sin(x+y) &= \sin(x)\cos(y) + \cos(x)\sin(y) \\ \cos(x+y) &=\cos(x)\cos(y) - sin(x)\sin(y) \\ \tan \theta &= \frac{\sin \theta}{\cos \theta}\end{aligned}$$
so the DE becomes:
$$-10K\sin(10t) \cos(\phi) - 10k \cos(10t) \sin(\phi) \\+ 2k \cos(10t) \cos(\phi) - 2k \sin(10t) \sin(\phi) = \cos(10t)$$
which i factor as:
$$\cos(10t)k(-10 \sin(\phi)  +  \cos(\phi)) \\+ \sin(10t) k(-10 \cos(\phi) - 2 \sin(\phi)) \\= \cos(10t)$$
here I can see that LHS has a $\sin(10t)$ that i need to be zero in order for the LHS to equal the RHS.  so that means that following needs to be zero:
$$-10 \cos(\phi) - 2 \sin(\phi) = 0$$
solving for $\phi$ i get:
$$\phi = tan^{-1}(-5)=-78.69^{\circ}$$
now i can plug $\phi$ in to equation and solve for k:
$$k = \frac{1}{10.2}$$
and that means my particular solution is:
$$y_p = \frac{1}{10.2}cos(10t - 78.69^{\circ})$$
The part i don't get is that when I solve this DE on wolfram alpha I get the particular solution of:
$$y_p = \frac{1}{52}\cos(10t) + \frac{5}{52}\sin(10t)$$
I'm wondering... why does my particular solution have only one sinusoidal term, but wolfram alpha has both a sin and cos sinusoidal term?  Also, it seems that I also have a phase and they don't have a phase...
How can i solve the DE to get this form?
$$y_p = \frac{1}{52}\cos(10t) + \frac{5}{52}\sin(10t)$$
 A: guessing that the particular solution is: 
$$y_p = A \cos(10t) + B \sin(10t)$$
$$y_p' = -10 A \sin(10t) + 10 B \cos(10t)$$
Thus:
$$y_p' + 2 y_p = \cos(10t)$$
$$(-10A \sin(10t) + 10B \cos(10t)) + 2(A \cos(10t) + B \sin(10t) = \cos(10t)$$
$$\cos(10t)(10B + 2A) + \sin(10t) (2B - 10A) = \cos(10t)$$
In order for LHS to equal RHS:
$$10B+2A = 1\tag{eq1}$$
$$2B - 10A = 0\tag{eq2}$$
solving for A and B:
$$A = \frac{1}{52}$$
$$B = \frac{5}{52}$$
Thus, the particular solution is:
$$y_p = \frac{1}{52} \cos(10t) + \frac{5}{52} \sin(10t)$$
A: The solution written by the OP is correct and finds the particular solution through the method of undetermined coefficients.
As the ordinary differential equation
$$\frac{dy}{dt}+2y=\cos(10t)$$
is a first order differential equation which is linear, we could also apply the integration factor technique. The general form of a first order linear equation is
$$\frac{dy}{dt}+p(t)y=g(t)$$
where both $p(t)$ and $g(t)$ are continuous functions. The integrating factor, $\mu(t)$, is given by 
$$\mu(t)=\large{e^{\int p(t)dt}}=e^{\int 2\,dt}=e^{2t}$$
therefore if we multiply every term by the integrating factor
$$e^{2t}\frac{dy}{dt}+2e^{2t}y=e^{2t}\cos(10t)$$
and rewrite the left-hand side of the equation by the product rule
$$\frac{d}{dt}(e^{2t}y)=e^{2t}\cos(10t)$$
we find
$$e^{2t}y=\int e^{2t}\cos(10t) \,dt\tag{*}$$
where we label the integral on the right-hand side as
$$I=\int e^{2t}\cos(10t) \,dt$$
and integrate by parts twice. In the first integration by parts, $u=\cos(10t)$ and $dv=e^{2t}\, dt$. Therefore
$$I=uv -\int v \,du=\frac{1}{2}e^{2t}\cos(10t)+5\int e^{2t}\sin(10t)\,dt$$
Next, take $u=\sin(10t)$ and $dv=e^{2t}\, dt$
$$I=\frac{1}{2}e^{2t}\cos(10t)+5\left(\frac{1}{2}e^{2t}\sin(10t)-5 \int e^{2t}\cos(10t)\,dt\right)$$
$$I=\frac{1}{2}e^{2t}\cos(10t)+\frac{5}{2}e^{2t}\sin(10t)-25I$$
we solve for $I$ to find
$$I=\frac{1}{52}e^{2t}\cos(10t)+\frac{5}{52}e^{2t}\sin(10t)+C$$
hence $(*)$ becomes
$$e^{2t}y=\frac{1}{52}e^{2t}\cos(10t)+\frac{5}{52}e^{2t}\sin(10t)+C$$
which after dividing by $e^{2t}$ forms the general solution
$$y(t)=\frac{1}{52}\cos(10t)+\frac{5}{52}\sin(10t)+Ce^{-2t}$$
The general solution is the same solution found by the method of undetermined coefficients. By finding the particular solution, the OP could then find the homogeneous solution and conclude
$$y_g(t)=y_h(t)+y_p(t)=Ce^{-2t}+\Big(\frac{1}{52}\cos(10t)+\frac{5}{52}\sin(10t)\Big)$$
