### Differential equations: Separation of variables

### Differential forms and separated variables

We have seen that finding an antiderivative of the function #f(t)# can be formulated as solving the first order differential equation #y'=f(t)#. An effective intermediate step from the ODE to the solution turned out to be its *differential form*: #\dd y = f(t)\,\dd t#. There are more first-order ODEs that can be solved using differential forms.

Separation of variables

Suppose that \(f(t)\) and \(g(y)\) are continuous functions, that #g(y)# is not the constant function #0#, that \(F(t)\) is an anti-derivative of #f(t)#, and that \(H(y)\) an anti-derivative of #\frac{1}{g(y)}#.

The general solution #y# of the differential equation \[y'(t)={g(y)}\cdot {f(t)}\] satisfies the equality \[H(y)=F(t)+C\] where \(C\) is a constant.

In general, this is not a solution of the ODE in the explicit form \( y(t)=\text{a function of }t\), but a relationship between the variables \(y\) and \(t\). Sometimes you can derive an explicit solution from this relationship.

Separable differential equation

Let #y# be a differentiable function of #t#. A **separable differential equation** is a first order differential equation of first degree in which the derivative \(y'\) can be factored as a product of a function \(g(y)\) of the unknown dependent variable \(y\) and a function \(f(t)\) of the independent variable \(t\): \[ y'(t) =g(y) \cdot f(t) \]

A solution such as #H(y)=F(t)+C# of the above theorem is often referred to as an **implicit solution** of the ODE.

If there are initial conditions, the implicit solution can be used for finding corresponding values of #C#.

Below are some examples. *Later* we will discuss the method in greater detail and provide more examples.

#y= \dfrac{1}{3}\; t + 1#

The differential equation is separable since, putting \(g(y) = \frac{1}{3} \) and \(f(t) = 1 \), we can rewrite it as

\[ 3\; \dd y = 1\; \dd t \]

After computing antiderivatives we have

\[ 3 y = t + C \]

Multiplying both sides by \(\frac{1}{3}\), we find

\[ y = \dfrac{1}{3}\; t + C \]

This is the *general solution* of the differential equation.

This solution denotes a class of functions, all of which have a rate of growth one third. Geometrically, we can represent the class of functions with the following *slope field.*

To obtain a *specific solution* we proceed to solve *initial value problem* with the initial condition \(\rv{t,y} = \rv{0,1}\). After substituting the values \(y=1\) and \(t=0\) we find

\[ 1 = \dfrac{1}{3}\; \cdot 0 + C \]

Therefore we have \(C = 1\), so the corresponding specific solution is the function:

\[ y = \dfrac{1}{3}\; t + 1 \]

This is represented graphically as a line inside the slope field.

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