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30 January 2004

A Trickier Separation

by George E. Hrabovsky, President, MAST

What We Did Last Time

We introduced the fundamental theorem of calculus and explored the method of separation of variables.

A Note of Apology

In the equation from last time I expressed the constant of integration for y as its initial value. This only occurs when x is zero. The solution to the equation will then be,

y   = x^3/3 + x_0 -=20= y_0 = x^3/3   - y_0 .

I should have added this, but I didn't, and so I apologize.

Where We Will Go This Time

I intend to develop a more significant example of separationof variables.

Separation of Variables, Part Two: An Introduction to Dummy Variables

Last time we solved a differential equation that was quite simple.  This time we look at an equation where each variable appears on one side of the equation,

FormBox[RowBox[{=C3=B7M^y=C3=B7L= y/=C3=B7Lx - (x + x^2), , =, , 0.}], TraditionalForm]

We separate this equation before we can integrate it. We canseparate the variables by moving the (x + x^2= ) term to the right hand side of the equation,

=C3=B7M^y=C3=B7Ly/=C3=B7Lx = (= x + x^2) .

This can be rewritten by considering y to be a function of x , y = y(x) ,

=C3=B7L/=C3=B7Lx=C3=B7M^y(x)=20= = (x + x^2) .

We can separate the variables totally,

=C3=B7M^y(x) = (x + x^2) =C3= =B7Lx .

We integrate the right hand side to get the expression

You will notice that each occurrence of x in the formula has been converted to x ' .  This is because we will be substituting in the limits of integration, as you will see below. The x ' symbols are placeholders and are called dummy variables .

Now comes the substitution,

Thus,

=C3=B7M^y(x) = (x^2/2 + x^3/3)= -    (x_0^2/2 + x_0^3/3) .

We solve this by taking the natural logarithm of both sides,

y(x) = ln[ (x^2/2 + x^3/3) - &= nbsp;  (x_0^2/2 + x_0^3/3)] .

This is the general solution to the differential equation.  By using specific initial values we can study specific solutions to the equation.

Mathematics Challenge from Last Time

The differential equation from last time was,

=C3=B7Ly/=C3=B7Lx = x^2 .

We can separate variables to get,

=C3=B7Ly = x^2=C3=B7Lx .

We can integrate the left hand side from some initial value of ycalled y_0 to y (and applying dummy variables) to get,

We can similarly integrate the right-hand side,

The equation is then,

y - y_0 = x^3/3,

or,

y = x^3/3 + y_0 .

You can see that I made an error in last week's column by not realizing that the sign of the initial value of =ywas wrong.  This form of integration removes the need for dealing with constants of integration.

Mathematics Challenge

Inserting initial values into the differential equation produces an initial-value problem. Speculate on how you would go about solving such a problem.

Created by Mathematica  (January 27, 2004)