(1) State whether the DE:

dy/dx = (2x + y^2)/ -2xy is exact

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

@N/@x = 2y

Since the two are equal, the DE is exact.

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

(3xy^4 + x)dx + (6x^2y^3 – 2y^2 + 7)dy = 0

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

M = (3xy^4 + x) and @M/@y = 12xy^3

N = (6x^2y^3 – 2y^2 + 7) and @N/@x = 12xy^3

So, exact.

Now, let:

f(x,y) = INT_x (3xy^4 + x) dx + C(y)

= 3x^2y^4/ 2 + x^2/ 2 + C(y)

Then: dC/dy = 6x^2y^3 -2y^2 + 7 – 6x^2y^3

= -2y^2 + 7

Therefore:

INT dC(y) = C(y) = INT (-2y^2 + 7)dy = -2y^3/ 3 + 7y

So the general solution is:

f(x,y) = 3x^2y^4/ 2 + x^2/ 2 -2y^3/ 3 + 7y

(3) Consider the differential equation:

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

Determine whether it is an exact DE or not. If it is, find the general solution and then the particular for an initial condition such that : y(1) = 3.

This is straightforward, having solved the previous example.

We have: M = (3x^2 y + 2) and @M/@y = 3x^2

And: N = (x^3 + y) and @N/@x = 3x^2

So, the DE is exact.

We have then:

f(x,y) = INT_x (3x^2 y + 2) dx + C(y) =

(x^3y + 2x) + C’(y)

INT dC(y)dy = C(y) = INT (x^3 + y)dy + c1= y^2/2 + c1

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

Or: x^3y + 2x + y^2/ 2 + c1 = c2 = INT_x (3x^2 y + 2) dx + C(y)

So the final general soln. is: : x^3y + 2x + y^2/ 2 + c = 0

To satisfy the condition y(1) = 3:

(1)^3 + 2(1) + (3)^2/ 2 + c = 0

So: c = -9 ½

Therefore: x^3y + 2x + y^2/2 = 19/2

**:**

*Integrating factors*The differential equation:

M(x,y) dx + N(x,y)dy = 0

Can always be transformed into an exact DE by multiplying it by some suitable factor, call it r(x,y). This makes the DE exact and is called an “integrating factor”.

*Usually*an appropriate r(x,y) can be found on inspection of the DE and visualizing how it might be most directly simplified, say if both sides were multiplied through by some expression.

Example: Find an integrating factor for the DE:

xdy + ydx = x^2y^2 dx and solve.

We can rewrite as:

xdy = [x^2y^2 –y]dx

and with M = x and N = [x^2y^2 –y], we easily see it’s

*not exact*.

Leaving the equation as is, one can see (if one is perceptive) that multiplying both sides by r(x,y) = 1/(x^2y^2) , will work wonders.

First, the right hand side simply becomes dx. E.g.

r(x,y)[xdy + ydx] = dx = 1/(x^2y^2)[ xdy + ydx]

Second, the savvy DE whiz kid (or calculus whiz kid) will quickly spot that the more complex side of the DE is easily reducible

*via the exact differential form**:

1/(x^2y^2)[ xdy + ydx] = d(- 1/xy)

Then our DE is quickly reduced to:

d(-1/xy) = dx

and integration yields:

INT d(-1/xy) = INT dx + c

For which we obtain:

-1/xy = x + c

Another more “refined” way to work with integrating factors starts with writing the typical first order linear DE as:

**dy/dx + Py = Q**

And the name of the game is to account for P and Q and also find the integrating factor, r.

Thus, one method for solving the DE shown is to find some function, usually r = r(x) such that if the equation is multiplied by r, the left side becomes the derivative of the product ry. That is:

r(dy/dx) + rPy = rQ

and we then make the effort to impose upon r the condition that:

r(dy/dx) + rPy = d/dx (ry)

which is not always easy, but often can be if one is clever enough!

Expanding the right side of the previous eqn. via differentials:

d/dx (ry) = (rdy + y dr) / dx

and adding to the left, gives:

r(dy/dx) + rPy + (-r (dy/dx) – y (dr/dx)) ->

dr/dx = rP

Then, if P = P(x) is a known function, we can solve for r:

Viz. dr/r = Pdx and ln r = INT Pdx + ln C

So: r = +/- Cexp(INT P dx)

And C can be taken as +/- C = 1

Then the function: r = exp(INT Pdx)

Is called the integrating factor

Example:

dy/dx + y = exp(x)

P = 1, Q = exp(x)

Then r = exp(INT dx) = exp(x)

So: exp(x)y = INT exp(2x) + C = exp(2x)/ 2 + c

And y = exp(x)/2 + Cexp(-x) or:

y = e^x/2 + Ce^-x

**Problems:**

(1) Solve: x^2y dy – xy^2 dx – x^3y^2dx = 0

(2) Solve using any method for integrating factors:

x (dy/dx) - 3y = x^2

* Tables of differential forms are available in most Calc texts.

Good luck!

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