# 2. Separation of Variables

### Later, on this page...

Some differential equations can be solved by the method of
**separation of variables** (or "variables separable") . This method is only possible if
we can write the differential equation in the form

A(x)dx+B(y)dy= 0,

where *A*(*x*) is a function of *x* only and
*B*(*y*) is a function of *y* only.

Once we can write it in the above form, all we do is integrate throughout, to obtain our general solution.

**NOTE:** In this variables separable section we only deal with **first order, first
degree** differential equations.

### Example 1 - Separation of Variables form

**a)** The differential equation (which we saw earlier in Solutions of Differential Equations):

`(dy)/(dx)ln\ x-y/x=0`

**can** be expressed in the required
form, *A*(*x*) *dx* + *B*(*y*) *dy* = 0, after some algebraic juggling:

` (dy)/(dx)ln\ x-y/x=0`

`dy\ ln\ x-(y\ dx)/x=0`

`dy-(y\ dx)/(x\ ln\ x)=0`

`(dy)/y-(dx)/(x\ ln\ x)=0`

`1/ydy-1/(x\ ln\ x)dx=0`

Here, `A(x) = -1/(x\ ln\ x)` and `B(y) = 1/y`.

[To solve the equation, we would then integrate throughout].

**b)** The following differential equation **cannot** be
expressed in the required form, so it cannot be solved using
separation of variables:

`(dy)/(dx)=(3(x+y))/(x(y-2))`

### Example 2

Solve the differential equation:

y^{2}dy+x^{3}dx =0

### Example 3

Solve the differential equation: `2(dy)/(dx)=(y(x+1))/x`

### Example 4

Solve `t(dx)/(dt)-x=3`

### Example 5

Solve** **`sqrt(1+4x^2)\ dy=y^3x\ dx`

Here is the graph of a typical solution for Example 5 where we have taken `K=50`:

## Particular Solutions

Our examples so far in this section have involved some constant of integration, *K*.

We now move on to see particular solutions, where we know some boundary conditions and we substitute those into our general solution to give a **particular solution**.

### Example 6

Find the particular solution for

`(dy)/(dx)+2y=6`

given that *x *= 0 when *y =* 1.

Here is the graph of our solution for Example 6:

### Example 7

Solve

`x\ dy = y\ ln\ y\ dx`

,

given

`x = 2` when `y = e`.

Here is the graph of our solution for Example 7:

### Example 8 - RL Circuit Application

In an RL circuit, the differential equation formed using Kirchhoff's law, is

`Ri+L(di)/(dt)=V`

Solve this DE, using separation of variables, given that

R= 10 Ω,L= 3 H andV= 50 volts, andi(0) = 0.

### Example 9 - Skydiver's Terminal Velocity

Image source.

When a skydiver jumps out of a (perfectly good) aeroplane, apart from experiencing sheer terror (and exhilaration), she experiences two forces:

Gravity, acting downwards

Air resistance, acting upwards

Gravity is written *g* and on earth its acceleration has value 9.8 ms^{-2}.

Air resistance depends on the size and shape of the item that is dropping through the air. We use the constant *k* to represent the amount of air resistance. It is called the **coefficient of drag**.

The air resistance is proportional to the **square of the velocity**, so the upwards resistance force due to air resistance can be represented by

F_{air}=kv^{2}.

Now the force on the object due to gravity is:

F_{gravity}=mg, wheregis the acceleration due to gravity.

The total force acting on the body is

`F_("total") = F_("air") - F_("gravity") = kv^2 - mg`

Dividing throughout by *m* gives us a good model for the velocity *v* of an object falling through the air:

`a=k/mv^2-g`

We can write this as

`(dv)/(dt)=k/mv^2-g`

Image source.

**Questions**

a. Find an expression for the velocity of a sky diver at time *t*.

b. Find the velocity after 5 seconds for a sky diver of mass *m* = 80 kg and *k* = 0.2

c. Find the terminal velocity for any object for general values of *g* and *k*.

d. Find the terminal velocity our skydiver with mass *m* = 80 kg and *k* = 0.2.

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