Solving separable differential equations

Form of a separable differential equation

We saw that first order linear equations are differential equations in the form

$\frac{dy}{dx}+P(x)y=Q(x)$

In contrast, first order separable differential equations are equations in the form

$N(y)\frac{dy}{dx}=M(x)$

$N(y)y'=M(x)$

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We call these “separable” equations because we can separate the variables onto opposite sides of the equation. In other words, we can put the $x$ terms on the right and the $y$ terms on the left, or vice versa, with no mixing.

Notice that the two previous equations above representing a separable differential equation are identical, except for the derivative term, which is represented by $dy/dx$ in the first equation and by $y'$ in the second equation.

When we’re working with separable differential equations, we usually prefer the $dy/dx$ notation (Leibniz notation), because this notation makes it easier to see how to separate variables. Starting with $dy/dx$, we move the $dx$ notation to the right, leaving just the $dy$ notation on the left:

$N(y)\ dy=M(x)\ dx$

How to solve separable equations

Most often, we’ll be given equations in the form

$N(y)\frac{dy}{dx}=M(x)$

Here’s the standard process we’ll follow to find a solution to this separable differential equation. First, we’ll separate variables by moving the $dx$ to the right side.

$N(y)\ dy=M(x)\ dx$

With variables separated, we’ll integrate both sides of the equation.

$\int N(y)\ dy=\int M(x)\ dx$

Normally when we evaluate an indefinite integral, we need to add in a constant of integration. So after integrating, we would expect to have something like

$\bold{N}(y)+C_1=\bold{M}(x)+C_2$

where $\bold{N}(y)$ is the integral of $N(y)$, and $\bold{M}(x)$ is the integral of $M(x)$. But when we solve for $\bold{N}(y)$, we get

$\bold{N}(y)=\bold{M}(x)+C_2-C_1$

What we have to realize here is that, if $C_2$ and $C_1$ are constants, then $C_2-C_1$ is also a constant. So we can make a substitution and represent $C_2-C_1$ as just one simple constant $C$.

$\bold{N}(y)=\bold{M}(x)+C$

For this reason, we don’t need to bother adding constants to both sides of the equation when we integrate. Instead of integrating to $\bold{N}(y)+C_1=\bold{M}(x)+C_2$, we can always skip these intermediate steps and go straight to $\bold{N}(y)=\bold{M}(x)+C$.

Once we have the equation in this form, the goal will be to express $y$ explicitly as a function of $x$, meaning that our solution equation is solved for $y$, like $y=f(x)$. Sometimes we won’t be able to get $y$ by itself on one side of the equation, and that’s okay. If we can’t, we’ll just settle for an implicit function, which means that $y$ and $x$ aren’t separated onto opposite sides of the solution.

In summary, our solution steps will be:

1.  If necessary, rewrite the equation in Leibniz notation.

2.  Separate variables with $y$ terms on the left and $x$ terms on the right

3.  Integrate both sides of the equation, adding $C$ to the right side

4.  If possible, solve the solution equation specifically for $y$.

Four steps to solve every separable differential equation

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Finding the solution of a separable differential equation

Let’s work through an example so that we can see these steps in action.

Example

Find the solution of the separable differential equation.

$y'=y^2\sin{x}$

Let’s write the equation in Leibniz notation, changing $y'$ to $dy/dx$.

$\frac{dy}{dx}=y^2\sin{x}$

Separate the variables, collecting $y$ terms on the left and $x$ terms on the right.

$dy=y^2\sin{x}\ dx$

$\frac{1}{y^2}\ dy=\sin{x}\ dx$

Sometimes we won’t be able to get y by itself on one side of the equation, and that’s okay.

With variables separated, and integrating both sides, we get

$\int \frac{1}{y^2}\ dy=\int \sin{x}\ dx$

$\int y^{-2}\ dy=\int \sin{x}\ dx$

$-y^{-1}=-\cos{x}+C$

$-\frac{1}{y}=-\cos{x}+C$

$\frac{1}{y}=\cos{x}+C$

Notice how we just multiplied through the equation by $-1$, but we didn’t change the sign on $C$. That’s because keeping $C$ is a little simpler than $-C$, and we’ll still end up with the same solution equation either way.

Finally, solving for $y$ gives the solution to the separable differential equation.

$1=y(\cos{x}+C)$

$y=\frac{1}{\cos{x}+C}$

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