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) = 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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