First, previous problems to solve:
(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
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
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
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
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
(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.