Trigonometric First Order Differential Equation I am trying to solve the following first order differential equation to find $v_c(t)$ as a function of time and frequency:
$RC \frac{dv_c(t)}{dt} + v_c(t) = \sin(2\pi ft)$
$R$ and $C$ are constants, $f$ being frequency and $t$ time
The assignment tells me the solution is of the general form:
$v_c(t) = a\cdot \sin(2\pi ft) + b \cdot \cos(2\pi ft)$
Here is what I did:
$\frac{dv_c(t)}{dt} = a \cdot (2\pi f) \cdot \cos(2\pi ft) - b \cdot (2\pi f) \cdot \sin(2\pi ft)$
And the whole equation ends up being this monstrosity:
$\sin(2\pi ft) = RC\;\Big(a (2\pi f) \cos(2\pi ft) - b (2\pi f) \sin(2\pi ft)\Big) \; + \;  a\cdot \sin(2\pi ft) + b \cdot \cos(2\pi ft)$
Question: how do I determine the coefficients $a$ and $b$ as a function of $f$ ?
Here is what the assignment suggests:

Set the amplitude values of the terms with $\sin(2\pi ft)$ and $\cos(2\pi ft)$ equal at both sides of the "$=$" sign, thus in the following way: 
  $g\sin(2\pi ft) + h\sin(2\pi ft)t = h\sin(2\pi ft)t$ 
  $m\cos(2\pi ft)t + n\cos(2\pi ft)t = q\cos(2\pi ft)$

I don't understand what the assignment is trying to make me do. Any help or tips are appriated, thank you.
Note: this is my first time touching on first order differential equations, sorry if the question is not formulated well.
 A: A small concept:
A Linear First Order Differential Equation is of the form:
$\frac {dy}{dx}+P(x)y=Q(x)$
Where $P(x)$ and $Q(x)$ are only functions of the independent variable (in this case $x$)
I do not want to go into the details of an integrating factor so, the general solution to such an equation is 
$ye^{\int P(x)dx}=\int Q(x)e^{\int P(x)dx}dx +C$, where $C$ is an arbitrary constant and the $e^{\int P(x)dx}$ is known as the integrating factor. (The integrating factor is simply a term that when multiplied with the equation at hand makes it an exact equation. Link is provided for details.)  
Solution:
First off, I am assuming here that $f$ is not a function of $t$. Observe that the equation becomes linear when both sides of the equation are divided by $RC$. The general solution in this case would be, 
$v_{c}e^\frac {t}{RC} =\frac {1}{RC}\int{e^\frac {t}{RC}sin(2\pi ft)}dt + B $
Where B is any arbitrary constant.  
$\Rightarrow v_{c}(t)e^\frac {t}{RC}=\frac {e^\frac {1}{RC}}{\frac {1}{(RC)^2} + (2\pi f)^2} (\frac {1}{(RC)^2} sin(2\pi ft)-\frac {2\pi f}{RC}cos(2\pi ft)) + B $
The value of the integral can be obtained by repeated application of integration by parts.
$\Rightarrow v_{c}(t)=\frac {1}{\frac {1}{(RC)^2} + {(2\pi f)^2}} (\frac {1}{(RC)^2} sin(2\pi ft)-\frac {2\pi f}{RC}cos(2\pi ft)) + B $
Given the general form, we simply equate the coefficients to get
$a=\frac {1}{1 + {(2\pi fRC)^2}}$ and $b=-\frac {1}{\frac {1}{RC2\pi f}+2\pi fRC}$
Hope this helps.
See this link https://en.wikipedia.org/wiki/Linear_differential_equation for more information on linear differential equations. See this link http://mathworld.wolfram.com/ExactFirst-OrderOrdinaryDifferentialEquation.html for information on exact equations of the First Order. Or get the book "Differential Equations" by S.L. Ross for a thorough course. 
A: $$\sin(2\pi ft) = RC\;\Big(a (2\pi f) \cos(2\pi ft) - b (2\pi f) \sin(2\pi ft)\Big) \; + \;  a\cdot \sin(2\pi ft) + b \cdot \cos(2\pi ft)$$
Rearrange terms in $\sin$ together and $\cos$ terms together too. 
$$\sin(2\pi ft)=\color{blue}{(a-bRC2\pi f)\sin(2\pi ft)} +\color{green}{(b+RCa2\pi f)\cos(2\pi ft)}$$
Now the coefficient of $\cos$ must be zero ...the coefficient of $\sin$ must be equal to 1
$$
\begin{cases}
(a-2bRC\pi f)=1\\
\\
(b+2RCa\pi f)=0\\
\end{cases}
$$
Solve the system for $a,b$ 
