(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.
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
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