solve differential equation $\frac{dP}{dt} = kPcos^{2}(rt-\Theta)$ I'm asked to solve the following differential equation.
$$\frac{dP}{dt} = kP \cos^{2}(rt-\theta)$$ 
$P(0) = P_{0} = 9$ for $k = 0.07, r = 0.49, \theta = 7.$
I've only done simple linear and separable differential equations up until this point, so I'm not sure how to approach this one. Any pointers or solutions are much appreciated.
 A: The equation
$\dfrac{dP}{dt} = kP \cos^2 (rt - \Theta) \tag 1$
is in fact of the variables-seperable type, to wit:
If $P(t') = 0$ for any $t' \in \Bbb R$, then by uniqueness of solutions $P(t) = 0$ for all $t \in \Bbb R$, since the zero solution satisfies $P(t') = 0$.  Furthermore, by if necessary reversing the sign of $P$ (a valid operation by linearity) if necessary we may take $P > 0$.  Therefore, for any solution which does not vanish identically we may write
$\dfrac{d\ln P}{dt} = \dfrac{1}{P}\dfrac{dP}{dt} = k \cos^2 (rt - \Theta); \tag 2$ 
we integrate 'twixt $t_0$ and $t$:
$\ln \dfrac{P(t)}{P(t_0)} = \ln P(t) - \ln P(t_0) = k \displaystyle \int_{t_0}^t \cos^2(rt - \Theta) \; dt; \tag 3$
we have in general (from a table of integrals):
$\displaystyle \int \cos^2 ax \; dx = \dfrac{x}{2} + \dfrac{\sin 2ax}{4a}; \tag 4$
setting
$\alpha = \dfrac{\Theta}{r}, \tag 5$
we write
$rt - \Theta = r(t - \alpha), \tag 6$
and
$\displaystyle \int_{t_0}^t \cos^2 (rt - \Theta) \; dt =  \int_{t_0}^t \cos^2 r(t - \alpha) \; dt = \left ( \dfrac{t - \alpha}{2} + \dfrac{\sin 2r(t - \alpha)}{4r}  \right \vert_{t_0}^t$
$= \dfrac{t - t_0}{2} + \dfrac{\sin 2r(t - \alpha) - \sin 2r(t_0 - \alpha)}{4r}$
$= \left ( \dfrac{t}{2} +  \dfrac{\sin 2r(t - \alpha)}{4r}\right ) - \left ( \dfrac{t_0}{2} +  \dfrac{\sin 2r(t_0 - \alpha)}{4r} \right ); \tag 7$
setting
$\beta(t_0) = \dfrac{t_0}{2} +  \dfrac{\sin 2r(t_0 - \alpha)}{4r}, \tag 8$
we write
$\displaystyle \int_{t_0}^t \cos^2 (rt - \Theta) \; dt =  \dfrac{t}{2} +  \dfrac{\sin 2r(t - \alpha)}{4r} - \beta(t_0); \tag 9$
returning to (3),
$\ln \dfrac{P(t)}{P(t_0)} = k \left ( \dfrac{t}{2} +  \dfrac{\sin 2r(t - \alpha)}{4r} \right ) - k\beta(t_0), \tag{10}$
or
$P(t) = P(t_0) \exp  \left (  k \left ( \dfrac{t}{2} +  \dfrac{\sin 2r(t - \alpha)}{4r} \right ) - k\beta(t_0) \right )$
$= P(t_0) e^{ - k\beta(t_0) } \exp  \left (  k \left ( \dfrac{t}{2} +  \dfrac{\sin 2r(t - \alpha)}{4r} \right )\right ); \tag{11}$
there is not much to be gained at this point by furhter re-arrangement of (11).  The reader my substitute specific values of the constants (including $t_0$ and $P(t_0)$) to realize a speciic, concrete solution.
