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.

Definition:

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) = ò_x (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.