1
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ Do you not think it's linear? Do you not think it's separable? $\endgroup$ – David Dec 10 '18 at 2:55
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.