## Saturday, July 20, 2013

### Exact Differential Equations (Problem Solutions)

The Problems Again:

(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.

Solutions:

1) We first re-arrange to obtain:

(-2xy)dy = (2x + y2) dx

And: (2x + y2) dx +(2xy)dy = 0

Then: M = (2x + y2) and N = (2xy)

Take partial derivatives:   M/y = 2y  and N/x = 2y

Since the two partials are equal, the DE is exact.

2) First, take the partials to make sure it’s exact:

M = (3xy4 + x)  and M/y = 12xy3

N = (6x2y3 – 2y2 + 7)  and N/x = 12xy3

So, it’s exact.  Now, let:

f(x,y) = òx (3xy4 + x) dx + C(y) = 3x2y4/ 2 + x2/ 2 + C(y)

Then: dC/dy = 6x2y3 -2y2 + 7 – 6x2y3  = -2y2 + 7

And:  ò dC(y) = C(y) = ò (-2y2 + 7) dy = -2y3/ 3 + 7y

So the general solution is:  f(x,y) = 3x2y4/ 2 + x2/ 2 -2y3/ 3 + 7y

3) This is straightforward, having solved the previous example.

We have: M = (3x2 y + 2)  so  M/y = 3x2

And: N = (x3 + y)   so  N/x= 3x2

So, the DE is exact. We have then:

f(x,y) = òx  (3x2 y + 2)   dx + C(y) = (x3 y + 2x) + C’(y)

whence:

òy dC(y)dy = C(y) = òy (x3 + y) dy + c1= y2/2 + c1

This solution satisfies: f(x,y) = c2

Or: x3y + 2x + y2/ 2 + c1 = c2 = òx (3x2 y + 2) dx + C(y)

So the final general soln. is: : x3y + 2x + y2/ 2 + c = 0

To satisfy the condition y(1) = 3:  (1)3 + 2(1) + (3)2/ 2 + c = 0

= 3 + 9/2 + c = 0

So: c = -7 ½

Therefore: : x3y + 2x + y2/ 2  - 15/2 = 0