(1) State whether the DE: dy/dx = (2x + y

^{2})/ -2xy is exact

(2) Show that the DE below is exact and find the solution

(3xy

^{4}+ x)dx + (6x

^{2}y

^{3}– 2y

^{2}+ 7)dy = 0

(3) Consider the differential equation:

(3x

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.

^{2}y + 2) dx + (x^{3}+ y) dy = 0Determine 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 + y

^{2}) dx

And: (2x + y

^{2}) dx +(2xy)dy = 0

Then: M = (2x + y

^{2}) 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 = (3xy

^{4}+ x) and ¶M/¶y = 12xy

^{3}

N = (6x

^{2}y

^{3}– 2y

^{2}+ 7) and ¶N/¶x = 12xy

^{3}

So, it’s exact. Now, let:

f(x,y) = ò

_{x}(3xy

^{4}+ x) dx + C(y) = 3x

^{2}y

^{4}/ 2 + x

^{2}/ 2 + C(y)

Then: dC/dy = 6x

^{2}y

^{3}-2y

^{2}+ 7 – 6x

^{2}y

^{3}= -2y

^{2}+ 7

And: ò dC(y) = C(y) = ò (-2y

^{2}+ 7) dy = -2y

^{3}/ 3 + 7y

So the general solution is: f(x,y) = 3x

^{2}y

^{4}/ 2 + x

^{2}/ 2 -2y

^{3}/ 3 + 7y

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

We have: M = (3x

^{2}y + 2) so ¶M/¶y = 3x

^{2}

And: N = (x

^{3}+ y) so ¶N/¶x= 3x

^{2}

So, the DE is exact. We have then:

f(x,y) = ò

_{x }(3x

^{2}y + 2) dx + C(y) = (x

^{3 }y + 2x) + C’(y)

whence:

ò

_{y}dC(y)dy = C(y) = ò

_{y}(x

^{3}+ y) dy + c1= y

^{2}/2 + c1

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

Or: x

^{3}y + 2x + y

^{2}/ 2 + c1 = c2 = ò

_{x}(3x

^{2}y + 2) dx + C(y)

So the final general soln. is: : x

^{3}y + 2x + y

^{2}/ 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: : x

^{3}y + 2x + y

^{2}/ 2 - 15/2 = 0

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