Thursday, July 18, 2013

Introduction to Partial derivatives- Exact Differential Equations

We have to introduce a bit about partial differentiation here, to make sense of what we call exact differential equations.


Given some differential equation: M(x,y)dx + N(x,y)dy = 0

If there exists a function f(x,y) such that:

f/x = M(x,y) and f/y = N(x,y)

Then the differential equation is said to be exact. (where f/x and f/y are the partial derivatives of f with respect to x and y, respectively)

A necessary and sufficient condition that the DE is exact is also that:

M/y = N/x

As an example, we want to find out if: dy/dx = (x + y)/ xy2 is exact

Re-arrange to get: (x + y)dx – xy2dy = 0

Then: M(x,y) = (x + y)

N(x, y) = (-xy2)

(Note: In the case of taking the partial derivative of an expression, say  M/y , one takes the derivative with respect to the variable in the denominator while holding the other constant.  Thus, the partial derivative with respect to x is taken by differentiating with respect to x, and holding y constant. Similarly, the partial derivative with respect to y is taken by differentiating with respect to y, and holding x constant. )

Thus: M/y = x and N/x = -2yx

So this DE is not exact since:  M/y  ¹  N/x

Example (2) Show that the DE:

2xydx + (1 + x2) dy = 0

is exact and find the general solution

As before: M(x,y) = 2xy and N(x.y) = (1 + x2)

Then: M/y = 2x and N/x = 2x

So yes, the DE is exact!

To obtain the general solution, let:

f(x,y) = òx (2xy dx + c(y)) = x2y + c(y)

since f /y = N we have:

/y [ x2y + c(y)] = x2 + d/dy [c(y)] = 1 + x2

(Since: /y [ x2y + c(y)] = x2 and c(y)/y = 0)

We see: d/dy [c(y)] = 1 and c(y) = y (e.g.  òdc(y) = c(y) = ò dy = y)

The function (or general solution) is then:

f(x,y) = x2y + c(y) = x2y + y or x2y +y = c

Problems for Math Mavens::

(1) State whether the DE: dy/dx = (2x + y2)/ -2xy is exact

(2) Show that the DE below is exact and find the solution:

(3xy4 + x)dx + (6x2y3 – 2y2 + 7)dy = 0

(3) Consider the differential equation:

(3x2 y + 2) dx + (x3 + y) dy = 0

Determine whether it is an exact DE or not. If it is, find the general solution and then the particular soln. for an initial condition such that : y(1) = 3.

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