At this point, you've learned how to estimate solutions to differential equations using direction fields and Euler's Method. You have used these methods because differential equations are often impossible to solve directly.
However, separable differential equations are a specific type of differential equation that can be solved explicitly, making them as special as sliced bread!
Meaning of Separable Differential Equations
Let's start by defining what exactly a separable differential equation is.
A first order separable differential equation is an equation that can be written in the form
\[y'=f(x)g(y).\]
Separable refers whether or not you can separate the \(x\) terms from the \(y\) terms. In general they can be separated into a function of \(x\) multiplied by a function of \(y\).
Identify the following first order equations as separable or non-separable.
(a) \(y'=(x^{2}+9)5y \)
(b) \(y'=3x^{2}-4x \)
(c) \(\ln(y') = x+5\)
(d) \(y'=xy+3x-2y-6 \)
(e) \(e^{y'} = x + y\)
Answer:
(a) If you let \(f(x) = x^{2}+9 \) and \(g(y) = 5y \), then you can write \(y' = f(x)g(y)\), so this is a separable equation.
(b) This one is already separated, although it might take you a moment to notice that you can let \(g(y) = 1\).
There is no reason that you can't let \(f(x)\) or \(g(y)\) be constant functions!
(c) The equation might not look like it is separable, but let's use properties of logarithms. Remember that \(\ln(y') = x+5\) means the same thing as
\[ y' = e^{x+5}.\]
So this is a separable equation.
(d) The differential equation \(y'=xy+3x-2y-6 \) doesn't look separable at first glance, but let's do some algebra just to be sure. If you factor the right hand side, you get
\[ \begin{align} y' &= xy+3x-2y-6 \\ \\ &= x(y+3) - 2(y+3) \\ \\ &= (x-2)(y+3). \end{align}\]
So in fact this equation is separable.
(e) Let's try using properties of logarithms on this example. If you do, \(e^{y'} = x + y\) becomes
\[ \ln e^{y'} = \ln (x+y), \]
or in other words
\[ y' = \ln (x+y) .\]
There is no way to write the right-hand side of that equation as \(f(x)g(y)\) because you have the \(x+y\) term inside the logarithm. Therefore this equation is not separable.
Separable Differential Equations
Separable differential equations can be used to model situations in a variety of disciplines. One such application is the mixing of some solution in a tank or vessel with another substance, such as salt. A solution of certain concentration enters a tank at a fixed rate. The mixture in the tank is thoroughly stirred. Then, it exits the tank at a fixed rate. The model of this problem results in a separable differential equation.
A real-world application of a "mixing problem" is the injection of a medication into the bloodstream. In this case, the medicine enters the bloodstream at a fixed rate. The medicine mixes into the bloodstream and flows through the body towards the heart. Once the medicine reaches the heart, the heart pumps the medicine in the bloodstream to the rest of your body at a fixed rate.
Applications of Separable Differential Equations
As mentioned at the beginning of the article, mixing problems are a common application of separable differential equations. Mixing problems can model anything from how the mixing of various chemicals and greenhouse gasses can affect the atmosphere to how beer is brewed.
Separable differential equations can also be used in Economics. We can use these equations to measure investments and how interest is compounded.
Newton's Law of Cooling is one of the more famous ways that separable equations are used. You can see many more ways that separable differential equations are used in the article Application of Separation of Variables.
Solving Separable Differential Equations
Now that you know what a separable differential equation is, and what they are used for, let's look at how to solve them. Suppose you have a separable differential equation that looks like this:
\[y'=f(x)g(y).\]
There are two cases to consider.
Case 1: If \(g(y) = 0\) for some value of \(y\).
Suppose that \(g(y) = 0\). Then you have \(y' = 0\).
The functions that give you \(0\) when you differentiate them are constant functions, so \(g(y) = 0\) corresponds to constant solutions to the separable differential equation.
In fact, if \(y_1, y_2, \dots , y_n\) are all roots of the equation \(g(y) = 0\), then the constant solutions are given by \(y = y_1\), \(y= y_2\), \(\dots\), \(y = y_n\).
It is always a good idea to look for constant solutions first.
Let's look at a quick example.
For the differential equation is \(y' = (x-2)(y+3) \), find any constant solutions.
Answer:
Constant solutions happen when \(g(y) = 0\). For this problem, \(g(y) = y + 3\), and this equals zero when \(y = -3\). So there is a constant solution, and it is \(y=-3\).
Case 2: For any value when \(g(y) \ne 0\).
Since \(g(y) \ne 0\), you can divide by it, giving you the equation
\[ \frac{1}{g(y)}y' = f(x).\]
If you let \(h(y) = 1/g(y)\), you can rewrite that a bit as
\[ h(y) y' = f(x).\]
Now let's integrate both sides with respect to \(x\), giving you
\[ \int h(y) y'(x) \,\mathrm{d}x = \int f(x)\,\mathrm{d}x ,\]
where it is made explicit that \(y\) is a function of \(x\). This leads you to the \(u\) substitution
\[ \begin{align} &u = y(x) \\ &\mathrm{d}u = y'(x) \mathrm{d}x . \end{align} \]
So the integral becomes
\[ \int h(u) \,\mathrm{d}u = \int f(x)\,\mathrm{d}x ,\]
and at this point you can hopefully integrate both sides and then back substitute to find the answer!
Let's look at some examples to see how it is done.
Examples of Separable Differential Equations
Solve the differential equation
\[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{3x^{2}}{\cos y}.\]
Answer:
Always check first to see that the equation is separable!
To make sure the differential equation is separable, you have to write the equation in the form \(y'=f(x)g(y)\).
If you let \(f(x)=3x^{2}\) and \(g(y)=1/ \cos y\), then you can see that the equation is, in fact, separable.
It is always a good idea to look for places where \(g(y) = 0\) since these correspond to constant solutions. In this case, \(g(y)\) can never be zero, though it can be undefined. So there won't be any constant solutions, but there might be places where the solution to the differential equation doesn't exist.
Next you want to set up the integral. Since
\[\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{3x^{2}}{\cos y},\]
or
\[\cos y \frac{\mathrm{d}y}{\mathrm{d}x} =3x^{2},\]
you might be tempted to "cross multiply by \(\mathrm{d}x \)" and then integrate. You can't actually do that because \(\mathrm{d}y/ \mathrm{d}x\) isn't really a fraction. But the really cool thing is that you can pretend to do it, and it makes the integral
\[ \int \cos y \,\mathrm{d}y = \int 3x^2 \,\mathrm{d}x ,\]
which is exactly what you would have gotten by doing the \(u\) substitution!
Then integrating you get
\[ \sin y = x^3 + C.\]
Remember that this is called an implicit solution, since you don't have \(y\) by itself. To get an explicit solution, you can solve for \(y\) and see that
\[ y(x) = \arcsin(x^3 + C).\]
What about an interval of existence? At the start of the problem, you noticed that when \(\cos y = 0\) there would be problems, and you can see from the explicit form of the solution that there is an arcsine in it, which is only defined on the interval
\[ \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right].\]
So that is the maximum interval of existence for the solution to the differential equation.
Let's take a look at another example.
Solve the differential equation \(y' = (x-2)(y+3) \).
Answer:
You have already checked in previous examples that this is a separable equation, and found that there is a constant solution \(y = -3\). So now let's find the rest of the solutions.
Setting up the integral,
\[ \int \frac{1}{y+3}\, \mathrm{d}y = \int x-2 \,\mathrm{d}x .\]
Then integrating gives you
\[ \ln|y+3| = \frac{1}{2}x^2 - 2x + C \]
as the implicit solution. Before finding an explicit solution, notice that you need \(y \ne -3\) for this solution to be valid, since otherwise you would be trying to take the logarithm of zero!
Going on to find the explicit solution, using properties of logarithms,
\[ |y+3| = \exp \left( \frac{1}{2}x^2 - 2x + C \right) \]
where \( \exp\) is just a handy notation for the exponential function.
Now let \(A = e^C\) since that is just another constant. Then you can write
\[ |y+3| = A\exp \left( \frac{1}{2}x^2 - 2x \right),\]
but for it to be a true explicit solution you need to finish solving for \(y\). In fact, this corresponds to two different solutions,
\[ y =-3+ A\exp \left( \frac{1}{2}x^2 - 2x \right),\]
and
\[ y =-3- A\exp \left( \frac{1}{2}x^2 - 2x \right).\]
You can combine these into one by using the \(\pm\) sign to indicate that it could be positive or negative, to get
\[ y =-3 \pm A\exp \left( \frac{1}{2}x^2 - 2x \right).\]
So does the explicit solution also give you the constant solution, or does that need to be written separately? Notice that if \(A=0\) then in fact you do get \(y=-3\), so the explicit solution also covers the constant solution. That means the solution to the differential equation is
\[ y =-3 \pm A\exp \left( \frac{1}{2}x^2 - 2x \right).\]
Separable Equations - Key takeaways
- A separable equation is an equation that can be written in the form \(y'=f(x)g(y)\).
- A separable differential equation can be separated into a function of \(x\) multiplied by a function of \(y\)
- The method behind solving separable differential equations involves moving all \(x\) and \(y\) variables onto their respective sides of the equation and integrating
- Separable differential equations have applications in finance, "mixing problems," and are used in Newton's Law of Cooling.
- Always remember to check for constant solutions and the interval of existence for solutions to separable differential equations.
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