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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