If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Differential equations>Unit 1

Lesson 7: Exact equations and integrating factors

# Integrating factors 1

Using an integrating factor to make a differential equation exact. Created by Sal Khan.

## Want to join the conversation?

• how can we know if (x + y) is not 0 and we can make the division?
• I'm also struggling with this question. If my reasoning below is not wrong, you have to consider this case separately, resulting in a new solution `x = 0`.

Lets recapitulate: we're looking for solutions for the equation `(3xy+y²) + (x²+xy)y' = 0`, that is, `<x, y>` pairs that satisfy the equation. The challenge to solve such equation is the differential part of it (the `y'`), so we find an equivalent equation `Ѱ(x,y) = C` without differentials, such any solution `<x, y>` of the last one should also be a solution of the former one.

In the next video (https://www.khanacademy.org/math/differential-equations/first-order-differential-equations/exact-equations/v/integrating-factors-2) we find that `Ѱ(x,y) = x³y + ½x²y²`, equivalent to `(3xy+y²) + (x²+xy)y' = 0` if and only if `x+y ≠ 0`.

Now lets see what happens for `x+y = 0`. The original equation
`(3xy + y²) + (x² + xy) y' = 0`,
turns into
`(-3x² + x²) + (x² - x²) y' = 0`,
that is,
`-2x² = 0`,
or simplified
`x = 0`.

That is, `x = 0` (the vertical Y-axis) is in fact a solution of our differential equation. But is this solution already included in our `Ѱ(x,y) = C` equation, that is, is `x = 0` a solution for it?

`Ѱ(0,y) = 0³y + ½0²y² = 0`

So `<0, y>` is a solution for the equivalent equation if and only if `C = 0`, although it is always a solution for the original equation.

To sum it up, the solutions for `(3xy+y²) + (x²+xy)y' = 0` are `Ѱ(x,y) = C` plus `x = 0`.

Please, don't read all of this without mistrust as I'm not a mathematician. I'm looking forward for corrections.
• How is Sal able to decide to use a function of x instead of y or y and x?
• As Sal said at you could even take a function of y or x and y. The choice is yours. As we are more familiar with a function of x, Sal decided to use a function of x.
• Why don't we apply the chain rule of differentiation when we r trying to prove that it is exact like the example here for M ? Shouldn't we apply the chain rule for 3xy ? and we did that only when we added the μ ?
• That's because we were taking the partial derivative with respect to y. If you take a partial derivative of 3xy, you treat x as a constant and you only take the derivative of y.
• Hey everyone, I am an 8th grader, and I love differential equations. I found this awesome document http://www.math.uah.edu/howell/DEtext/Part2/Exact.pdf .But I do not understand the solution to 7.5 a. My answer is (-y^3/x^2)+y^4=C. I found the integrating factor, which is x^-3, and I substituted it into the equation and solved. But when I punched it into wolfram alpha , it got a different solution of (y^3/x^2)-y^4=C. Which is correct?
(1 vote)
• The answers are both correct. Wolfram alpha took your answer and multiplied by -1. On the left hand side, C is any constant- so when they multiplied and got -C, it is still any constant- so they merely called it C. Since C could be any number before the multiplication, and could still be any number after the multiplication, it doesn't make sense to have a -C.
• How much is this playlist connected to the Calculus playlist?
• Differential Equations is a natural extension of calculus. Solving differential equations demands skill in many areas of Algebra and Calculus. In terms of connectedness with the Calculus playlist: it seems as though Sal treats the viewer as though they are comfortable with differential, integral, and multivariable calculus.
• What if (x+y) did not cancel on both sides, and we were left with a y term in u(x)?
• Hmm, it does not seem easy if you try to pick up a function /mu(y) instead of /mu(x).. Any ideas on that?
• try multiplying the original equation by dx/dy then considering x as a function of y before looking for an integrating factor
• In this video you say that it is possible to figure out the multiplier if it contains x and y. How is this possible?