### Differential equations: Separation of variables

### Solving ODEs by separation of variables

We describe the steps involved in solving a separable differential equation.

The separation of variables method

The steps in the *separation of variables,* applied to \(y '= g (y) \cdot f (t) \) , are the following:

- Transform the equation in differential form with all expressions containing \(y\) and all expressions containing \(t\) on different sides: \[\frac{1}{g(y)}\,\dd y = f(t)\,\dd t\]
- Integrate both sides of the equation in step 1 in order to obtain an implicit solution of the ODE
*:*\(H(y)=F(t)+C\) where \(C\) is a constant, \(F(t)\) is an anti-derivative of #f(t)# and \(H(y)\) is an anti-derivative of #\frac{1}{g(y)}#. - If an
*initial condition*is given, then use it to find a*special solution*for \(y\) from the*general solution*obtained in the previous step, or, should no explicit solution be found, from step 2. - Try to make explicit the solution as explicit as possible, by solving #y# from the equation of step 2 or 3.

We illustrate the separation method on the basis of some examples.

\(y(t)=K+C\cdot \e^{-r\cdot t}\)

In differential form the given differential equation is \[\dd y=r\cdot(K-y)\,\dd t\] The constant function #y(t)=K# is a solution which we put aside for the time being, so we can divide both sides by #K-y#. Separation of variables gives\[\frac{1}{K-y}\,\dd y=r\,\dd t\] The left and right side can be written as \(\dd\bigl(\ln(|K-y|)\bigr)\) and \(\dd(r\cdot t)\), respectively. As a consequence we have the following equality between two differentials: \[\dd\bigl(\ln(|K-y|)\bigr)=\dd(r\cdot t)\] This means the functions behind the d's are equal to each other up to a constant: \[\ln(|K-y|)=r\cdot t+c\] for a constant \(c\). Therefore, \[|y-K| =\e^{-r\cdot t+c}=\e^c\cdot \e^{-r\, t}=C\cdot \e^{-r\, t}\] for a certain \(C\gt 0\). Eliminating the absolute signs leads us to the explicit solution: \[y=K+C\cdot \e^{-r\, t}\] for some constant \(C\ne0\). Since for #C=0# we recover the solution #y=K# already found, we do not need to rule out this value of #C#. Thus we find that the general solution is \[y(t)=K+C\cdot \e^{-r\, t}\] where \(C\) is an arbitrary constant.

which, according to the

*theorem Solution of the exponential growth equation,*has the general solution \(u=D\cdot\e^{-r\cdot t}\) where #D# is the integration constant. Expressed in terms of #y#, this reads \[y=K-u =K- D\cdot\e^{-r\cdot t}\] from which the general solution described above emerges when we take #D=-C#.

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