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03. Separable Equations

Dated: 08-11-2024

The differential equation

\[\frac{dy}{dx} = f(x, y)\]

is called separable equation if it can be written as

\[\frac{dy}{dx} = h(x)g(y)\]

To solve separable equations, we perform the following steps

  • We find \(g(y) = 0\) to find constant solutions to the equation.
  • For non constant solutions, we write the equation as
\[\frac{dy}{g(y)}=h(x)dx\]

Then integrate1 it

\[\int \frac{1}{g(y)} dy = \int h(x) dx\]

To get the solution of the form

\[G(y) = H(x) + C\]
  • We list the entire constant and non-constant solutions to avoid repetition.
  • If you are given Initial value problem,2 use the initial condition to find the particular solution.
Note
  • No need to use 2 constants of integration1 because \(C_1 - C_2 = C\)
  • The constants of integration1 can be relabeled in a convenient way.
  • Since a particular solution may coincide with a constant solution, step 3 is important.

Example

\[\frac{dy}{dt}=1+t^{2}+y^{2}+t^{2}y^{2}, y(0)=1\]
\[\because 1+t^{2}+y^{2}+t^{2}y^{2}=(1+t^{2})(1+y^{2})\]

Step 1

\[1 + y^2 = 0\]
\[\implies y = \pm \sqrt {-1}\]

There are no real solutions

Step 2

\[\frac{dy}{1+y^{2}}=(1+t^{2})dt\]
\[\int \frac{dy}{1+y^{2}}=\int(1+t^{2})dt\]
\[\tan^{-1}(y)=t+\frac{t^{3}}{3}+C\]
\[y=\tan\left(t+\frac{t^{3}}{3}+C\right)\]

Step 3

Since there are no real solutions, all solutions are given by implicit and explicit solutions

Step 4

The initial condition \(y(0) = 1\) gives

\[C=\tan^{-1}(1)=\frac{\pi}{4}\]

Hence the solution to this initial value problem2 is

\[\tan^{-1}(y)=t+\frac{t^{3}}{3}+\frac{\pi}{4}\]
\[y=\tan\left(t+\frac{t^{3}}{3}+\frac{\pi}{4}\right)\]

Example

\[\frac{dy}{dx}=(y-1)^{2}, y(0)=1\]

Step 1

\[(y - 1)^2 = 0 \implies y = 1\]

Step 2

\[\frac{dy}{(y-1)^{2}}=dx\]
\[\int(y-1)^{-2}dy=\int dx\]
\[\frac{(y-1)^{-2+1}}{-2+1}=x+c\]
\[-\frac{1}{y-1}=x+c\]

Step 3

All solutions of the equation are

\[ = \begin{cases} -\frac{1}{y-1}=x+c \\ y = 1 \end{cases} \]

Step 4

\[\because y(0) = 1\]
\[- \frac 1 {y(x) - 1} = x + c\]
\[\because x \to 0\]
\[\implies y(x) - 1 \to 0\]
\[\implies - \frac 1 {y(x) - 1} \to -\infty\]
\[\implies c \to -\infty\]

Plug it back into the equation

\[- \frac 1 {y - 1} = - \infty\]
\[\frac 1 \infty = y - 1\]
\[\implies y = 1\]

Hence the solution is \(y = 1\).

References

Read more about notations and symbols.

  1. Read more about integration. ↩↩↩

  2. Read more about initial value problem. ↩↩