Separation of Variables (College Board AP® Calculus AB): Revision Note

Separation of variables

What is separation of variables?

• Separation of variables can be used to solve certain types of first order differential equations

• Look out for equations of the form $\frac{\mathit{d}y}{\mathit{d}x}=g\left(x\right)h\left(y\right)$

  • I.e. $\frac{\mathit{d}y}{\mathit{d}x}$ is equal to a function of $x$ multiplied by a function of $y$

  • Be careful – the ‘function of $x$’ $g\left(x\right)$ may just be a constant!

    • For example in $\frac{\mathit{d}y}{\mathit{d}x}=6y$, $g\left(x\right)=6$ and $h\left(y\right)=y$

• If the equation is in that form

  • then you can use separation of variables to try to solve it

How do I solve a differential equation using separation of variables?

• STEP 1
  Rearrange the equation into the form $\left(\frac{1}{h\left(y\right)}\right)dy=g\left(x\right)dx$

  • E.g. $\frac{dy}{dx}=-x{y}^3\Rightarrow -\frac{1}{y^3}dy=xdx$

    • $g\left(x\right)=x,h\left(y\right)=-y^3$

• STEP 2
  Integrate both sides with respect to $x$

  • This changes the equation into the form $\int \frac{1}{h\left(y\right)}dy=\int g\left(x\right)dx$

  • E.g.$-\frac{1}{y^3}dy=xdx\Rightarrow \int \frac{-1}{y^3}dy=\int xdx$

    • You can think of this step as ‘multiplying the $\mathit{d}x$ across and integrating both sides’

      • Mathematically that’s not quite what is happening, but it will get you the right answer here!

• STEP 3
  Work out the integrals on both sides of the equation

  • Don’t forget to include a constant of integration

    • You only need one constant of integration, even though there are two integrals

  • E.g. $\int \frac{-1}{y^3}dy=\int xdx\Rightarrow -\int y^{-3}dy=\int xdx\Rightarrow \frac{1}{2}{y}^{-2}=\frac{1}{2}{x}^2+C$

• STEP 4
  Rearrange the solution

  • E.g. $\frac{1}{2}{y}^{-2}=\frac{1}{2}{x}^2+C\Rightarrow \frac{1}{y^2}=x^2+2C\Rightarrow y^2=\frac{1}{x^2+2C}$

  • Note that you won't always be able to rewrite the solution in $y=f\left(x\right)$ form

    • In this case $y=\sqrt{\frac{1}{x^2+2C}}$ is not correct, because the solutions also include the $y=-\sqrt{\frac{1}{x^2+2C}}$ option

    • If an exam question requires the answer in a particular form, be sure to rearrange into that form

  • Also note that $2C$ is just another arbitrary integration constant

    • So $y^2=\frac{1}{x^2+C}$ would be a 'neater' way to write the solution

• This method gives the general solution to the differential equation

  • For finding the particular solution, see the 'Particular Solutions' study guide