Exact Equations and Integrating Factors

Example 1: Solve

(3x²y³ − 5x⁴) dx + (y + 3x³y²) dy = 0

In this case we have:

• M(x, y) = 3x²y³ − 5x⁴
• N(x, y) = y + 3x³y²

We evaluate the partial derivatives to check for exactness.

• ∂M∂y = 9x²y²
• ∂N∂x = 9x²y²

They are the same! So our equation is exact.

We can proceed.

Now we want to discover I(x, y)

Let's do the integration with x as an independent variable:

I(x, y) = ∫M(x, y) dx

= ∫(3x²y³ − 5x⁴) dx

= x³y³ − x⁵ + f(y)

Note: f(y) is our version of the constant of integration "C" because (due to the partial derivative) we had y as a fixed parameter that we know is really a variable.

So now we need to discover f(y)

At the very start of this page we said that N(x, y) can be replaced by ∂I∂y, so:

∂I∂y = N(x, y)

Which gets us:

3x³y² + dfdy = y + 3x³y²

Cancelling terms:

dfdy = y

Integrating both sides:

f(y) = y²2 + C

We have f(y). Now just put it in place:

I(x, y) = x³y³ − x⁵ + y²2 + C

and the general solution (as mentioned before this example) is:

I(x, y) = C

Ooops! That "C" can be a different value to the "C" just before. But they both mean "any constant", so let's call them C₁ and C₂ and then roll them into a new C below by saying C=C₁+C₂

So we get:

x³y³ − x⁵ + y²2 = C

And that's how this method works!

Example 2: Solve

(3x² − 2xy + 2)dx + (6y² − x² + 3)dy = 0

• M = 3x² − 2xy + 2
• N = 6y² − x² + 3

So:

• ∂M∂y = −2x
• ∂N∂x = −2x

The equation is exact!

Now we are going to find the function I(x, y)

This time let's try I(x, y) = ∫N(x, y)dy

So I(x, y) = ∫(6y² − x² + 3)dy

I(x, y) = 2y³ − x²y + 3y + g(x)    (equation 1)

Now we differentiate I(x, y) with respect to x and set that equal to M:

∂I∂x = M(x, y)

0 − 2xy + 0 + g'(x) = 3x² − 2xy + 2

−2xy + g'(x) = 3x² − 2xy + 2

g'(x) = 3x² + 2

And integration yields:

g(x) = x³ + 2x + C     (equation 2)

Now we can replace the g(x) in equation 2 in equation 1:

I(x, y) = 2y³ − x²y + 3y + x³ + 2x + C

And the general solution is of the form

I(x, y) = C

and so (remembering that the previous two "C"s are different constants that can be rolled into one by using C=C₁+C₂) we get:

2y³ − x²y + 3y + x³ + 2x = C

Solved!

Example 3: Solve

(xcos(y) − y)dx + (xsin(y) + x)dy = 0

We have:

M = (xcos(y) − y)dx

∂M∂y = −xsin(y) − 1

N = (xsin(y) + x)dy

∂N∂x = sin(y) +1

Thus

∂M∂y ≠ ∂N∂x

Example 4: Solve

[y² − x²sin(xy)]dy + [cos(xy) − xy sin(xy) + e^2x]dx = 0

M = cos(xy) − xy sin(xy) + e^2x

∂M∂y = −x²y cos(xy) − 2x sin(xy)

N = y² − x²sin(xy)

∂N∂x = −x²y cos(xy) − 2x sin(xy)

They are the same! So our equation is exact.

This time we will evaluate I(x, y) = ∫M(x, y)dx

I(x, y) = ∫(cos(xy) − xy sin(xy) + e^2x)dx

 Using Integration by Parts we get:

I(x, y) = 1ysin(xy) + x cos(xy) − 1ysin(xy) + 12e^2x + f(y)

I(x, y) = x cos(xy) + 12e^2x + f(y)

Now we evaluate the derivative with respect to y

∂I∂y = −x²sin(xy) + f'(y)

And that is equal to N, that equal to M: 

∂I∂y = N(x, y)

−x²sin(xy) + f'(y) = y² − x²sin(xy)

f'(y) = y² − x²sin(xy) + x²sin(xy)

f'(y) = y² 

f(y) = 13y³

So our general solution of I(x, y) = C becomes:

xcos(xy) + 12e^2x + 13y³ = C

Done!

Example 5: (y² + 3xy³)dx + (1 − xy)dy = 0

M = y² + 3xy³

∂M∂y = 2y + 9xy²

N = 1 − xy

∂N∂x = −y

So it's clear that ∂M∂y ≠ ∂N∂x

But we can try to make it exact by multiplying each part of the equation by x^myⁿ:

Which "simplifies" to:

(x^myⁿ⁺² + 3x^m+1yⁿ⁺³)dx + (x^myⁿ − x^m+1yⁿ⁺¹)dy = 0

And now we have:

M = x^myⁿ⁺² + 3x^m+1yⁿ⁺³

∂M∂y = (n + 2)x^myⁿ⁺¹ + 3(n + 3)x^m+1yⁿ⁺²

N = x^myⁿ − x^m+1yⁿ⁺¹

∂N∂x = mx^m−1yⁿ − (m + 1)x^myⁿ⁺¹ 

And we want ∂M∂y = ∂N∂x

So let's choose the right values of m and n to make the equation exact.

Set them equal:

(n + 2)x^myⁿ⁺¹ + 3(n + 3)x^m+1yⁿ⁺² = mx^m−1yⁿ − (m + 1)x^myⁿ⁺¹ 

Re-order and simplify:

[(m + 1) + (n + 2)]x^myⁿ⁺¹ + 3(n + 3)x^m+1yⁿ⁺² − mx^m−1yⁿ = 0 

For it to be equal to zero, every coefficient must be equal to zero, so:

1. (m + 1) + (n + 2) = 0
2. 3(n + 3) = 0
3. m = 0

That last one, m = 0, is a big help! With m=0 we can figure that n = −3

And the result is:

x^myⁿ = y⁻³

We now know to multiply our original differential equation by y⁻³:

Which becomes:

(y⁻¹ + 3x)dx + (y⁻³ − xy⁻²)dy = 0

And this new equation should be exact, but let's check again:

M = y⁻¹ + 3x

∂M∂y = −y⁻²

N = y⁻³ − xy⁻²

∂N∂x = −y⁻²

∂M∂y = ∂N∂x

They are the same! Our equation is now exact!

So let's continue:

I(x, y) = ∫N(x, y)dy

I(x, y) = ∫(y⁻³ − xy⁻²)dy

I(x, y) = −12y⁻² + xy⁻¹ + g(x)

Now, to determine the function g(x) we evaluate

∂I∂x = y⁻¹ + g'(x)

And that equals M = y⁻¹ + 3x, so:

y⁻¹ + g'(x) = y⁻¹ + 3x

And so:

g'(x) = 3x

g(x) = 32x²

So our general solution of I(x, y) = C is:

−12y⁻² + xy⁻¹ + 32x² = C

Example 6: (3xy − y²)dx + x(x − y)dy = 0

M = 3xy − y²

∂M∂y = 3x − 2y

N = x(x − y)

∂N∂x = 2x − y

∂M∂y ≠ ∂N∂x

So, our equation is not exact.

Let us work out Z(x):

So Z(x) is a function only of x, yay!

So our integrating factor is

u(x) = e^∫Z(x)dx

= e^∫(1/x)dx

= e^ln(x)

= x

Now that we found the integrating factor, let's multiply the differential equation by it.

x[(3xy − y²)dx + x(x − y)dy = 0]

and we get

(3x²y − xy²)dx + (x³ − x²y)dy = 0

It should now be exact. Let's test it:

M = 3x²y − xy²

∂M∂y = 3x² − 2xy

N = x³ − x²y

∂N∂x = 3x² − 2xy

∂M∂y = ∂N∂x

So our equation is exact!

Now we solve in the same way as the previous examples.

I(x, y) = ∫M(x, y)dx

= ∫(3x²y − xy²)dx

= x³y − 12x²y² + c₁

And we get the general solution I(x, y) = c :

x³y − 12x²y² + c₁ = c

Combine the constants:

x³y − 12x²y² = c

Solved!