# 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.

• Do you not think it's linear? Do you not think it's separable? – David Dec 10 '18 at 2:55

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.