# Integrating factor in Exact Equations

• MHB
• alane1994
In summary, you have a differential equation that you checked to be exact and found that it has the form, $$(3x^2y+2xy+y^3)+(x^2+y^2)y\prime=0$$. You then found that the integrating factor is in the form, $$\dfrac{M_y(x,y)-N_x(x,y)}{N(x,y)}\mu$$. You solved the ODE by separation of variables to get, $$C=(3x^2y+y^3)e^{3x}$$.
alane1994
Ok, so I have this differential equation.

$(3x^2y+2xy+y^3)+(x^2+y^2)y\prime=0$

First I needed to check to see if it is exact.

$$M=3x^2y+2xy+y^3$$
$$N=x^2+y^2$$

$$\dfrac{\partial M}{dy}(3x^2y+2xy+y^3)=3x^2+2x+3y^2$$

$$\dfrac{\partial N}{dx}(x^2+y^2)=2x+0$$

For the integrating factor, I believe it is in the form,

$\dfrac{M_y(x,y)-N_x(x,y)}{N(x,y)}\mu$

And so, I would have,

$$\dfrac{3x^2+2x+3y^2-2x}{x^2+y^2}\mu = \dfrac{3x^2+3y^2}{x^2+y^2}\mu=\dfrac{d\mu}{dx}$$

Now... I feel as though I am wrong in this. Could someone look and tell me if I am on the right track, and if I am off guide me back to the path of righteousness?

Last edited:
alane1994 said:
And so, I would have,

$$\dfrac{3x^2+2x+3y^2-2x}{x^2+y^2}\mu = \dfrac{3x^2+3y^2}{x^2+y^2}\mu=\dfrac{d\mu}{dx}$$

Now... I feel as though I am wrong in this. Could someone look and tell me if I am on the right track, and if I am off guide me back to the path of righteousness?
You seem to be on the right track, but you need to go a little bit further, by noticing that $\dfrac{3x^2+3y^2}{x^2+y^2} = \dfrac{3(x^2+y^2)}{x^2+y^2} = 3.$ (Happy)

$$\dfrac{d\mu}{dx}=3\mu$$

Would $$\mu(x)=0$$?

alane1994 said:
$$\dfrac{d\mu}{dx}=3\mu$$

Would $$\mu(x)=0$$?
No. Think again (think exponential).

alane1994 said:
$$\dfrac{d\mu}{dx}=3\mu$$Would $$\mu(x)=0$$?
Sure, but things would be boring if that was the only solution. Instead, try solving the ODE by separation of variables.

EDIT: Ninja'd by Opalg.

I feel foolish, I am unsure how to get what is required.
I typed it into wolfram and got,
$$\mu(x)=e^{3x}$$

$$\dfrac{d\mu}{dx}=3\mu$$
$$\dfrac{1}{\mu}\dfrac{d\mu}{dx}=3$$
$$\dfrac{1}{\mu}d\mu=3dx$$
Then solve yeah?

alane1994 said:
$$\dfrac{d\mu}{dx}=3\mu$$
$$\dfrac{1}{\mu}\dfrac{d\mu}{dx}=3$$
$$\dfrac{1}{\mu}d\mu=3dx$$
Then solve yeah?
Yes! (Smile)

$$C=(3x^2y+y^3)e^{3x}$$

EDIT: Is this correct?

Last edited:

## What is an integrating factor?

An integrating factor is a function that is used to transform an exact differential equation into an easier form for solving. It is typically denoted by the symbol "μ" and is multiplied to both sides of the equation to make it exact.

## Why do we need integrating factors?

Integrating factors are necessary because not all differential equations are exact. By multiplying an integrating factor to both sides of an equation, we can make it exact and therefore easier to solve.

## How do we determine the integrating factor for an exact equation?

The integrating factor for an exact equation can be determined by finding the function "μ" that satisfies the condition ∂(μM)/∂y = ∂(μN)/∂x. This is known as the integrating factor equation.

## Can any function be an integrating factor?

No, not all functions can be an integrating factor. The function must satisfy the integrating factor equation in order to be considered an integrating factor for a specific exact equation.

## Are there any other uses for integrating factors besides solving exact equations?

Yes, integrating factors can also be used for solving non-exact equations, finding exact solutions for first-order linear equations, and solving certain types of differential equations involving trigonometric functions.

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