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) = ò_x (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) = ò_x  (3x² y + 2)   dx + C(y) = (x³ y + 2x) + C’(y)


whence:

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


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

Or: x³y + 2x + y²/ 2 + c1 = c2 = ò_x (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