2. Separation of Variables

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:

y2 dy + x3 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`

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.

Example 7

Solve

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

given

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

Example 8 - RL Circuit Application

RL circuit diagram

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 and V = 50 volts, and i(0) = 0.

Example 9 - Skydiver's Terminal Velocity

skydiver leaping from Cessna
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

Fair = kv2.

Now the force on the object due to gravity is:

Fgravity = mg, where g is 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`

skydivers
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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