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Definition

A separable differential equation is one of the simplest types of differential equations. In basic terms, it is a differential equation where you can put the x’s on one side, and y’s on the other.

Solving

This is the general form of a differential equation. $$\dfrac{dy}{dx} = f(x, y)$$ A separable differential equation is one that can be written as $$\dfrac{dy}{dx} = f(x)g(y)$$ Letting $h(y) = \frac{1}{g(y)}$, we can write $$\int h(y)\, \mathrm{d}y = \int f(x)\, \mathrm{d}x$$

Example 1:

$$\dfrac{dy}{dx} = \dfrac{sec^2(y)}{1 + x^2}$$ $$\int cos^2(y)\, \mathrm{d}y = \int \dfrac{1}{1 + x^2}\, \mathrm{d}x$$ $$\cos(2y) = 2\cos^2(y) - 1 \to \cos^2(y) = \frac{\cos(2y) + 1}{2}$$ $$\int \frac{\cos(2y) + 1}{2}\, \mathrm{d}y = arctan(x) + C$$ $$\sin(2y) + y = 2\arctan(x) + C$$ From this point, it's hard to simplify, so you can just leave it in this form.

Example 2: Newton’s Law of Cooling

If you have a hot cup of coffee, it cools really quickly in the beginning. As the temperature of the coffee gets closer to room temperature, it starts cooling slower. Once the coffee is almost room temperature, it basically stays constant. If the coffee starts off at room temperature, it doesn’t cool at all. This is called Newton’s Law of Cooling. The rate of the temperature change is proportional to the difference of the body’s temperature and the surrounding’s temperature.

$$\dfrac{dT}{dt} = k(T-M)$$ Here, we assume that the temperature of the body, T, is greater than the temperature of the surroundings, M. Let's think of the sign of k. The rate will be negative, so the sign of k is negative. Let's just replace this with -k. $$\dfrac{1}{T-M}\,\mathrm{d}T = -k\,\mathrm{d}t$$ $$\int \dfrac{1}{T-M}\,\mathrm{d}T = \int -k\,\mathrm{d}t$$ $$\ln|T-M| = -kt + C$$ $T \geq M$, so we can take out the absolute value. $$\ln(T-M) = -kt + C$$ $$T-M = e^Ce^{-kt}$$ $$T = M + De^{-kt} \text{ (D is positive)}$$

Missing Solution

Without the condition that T is greater than M, we would be missing a solution. If T = M, then the constant D would be 0. Why did we get this issue? We divided by T-M in the beginning, which only works if that’s non-zero.

Practice Problem

$$y' = k(y - 60) \to y = 60 + Ce^{kt}$$ $$t = 0, y = 100 \to 100 = 60 + C\to C = 40$$ $$t = 10, y = 90 \to 90 = 60 + 40e^{10k}$$ $$\dfrac{3}{4} = e^{10k} \to k = \frac{ln(3/4)}{10} = -0.029$$ We have our final equation, with all of the coefficients assigned: $$\boxed{y = 60 + 40e^{-0.029t}}$$ Now, we just plug in 80 for y and we get t = 23.9 minutes.

Example 3: Radioactive Decay

For this case, the rate at which a radioactive isotope decays depends on the current amount of that isotope.

So, y’ = ky

Practice Problem

Pu-239's half life is 24,100 years. Now, suppose 10 grams of Pu-239 was released. How long will it take for the 10 grams to decay to 1 gram?

$$y = Ce^{kt}$$ This is the general form of our equation. Now, we plug in t = 0, and y = 10, to get C = 10. $$y = 10e^{kt}$$ Now, we plug in t = 24,100 years and amount = 5 grams. $$5 = 10e^{24100k} \to k = -0.00002876$$ $$1 = 10e^{-0.00002876t} \to t = \text{ 80,062 years}$$
David Witten