# Integrating Factor and Unintegratable Term

• Sierevello
In summary: ODE to solve and it seems simple enough. However, after you have used the integrating factor the integral is not integratable. He suggests using the method of Frobenius / power series method.
Sierevello
I was given the following ODE to solve and it seemed simple enough. However, after you have used the integrating factor the integral is not integratable.

y' = (1+x^2)y +x^3, y(0)=0

Find y(1) if y(x) is the solution to the above ODE.

So I put it in the proper form of:

y' + (-1-x^2)y = x^3

INT FACTOR = e^(INTEGRAL(-1-x^2) dx) = e^(-x^3/3 - x) so:

y = e^(x/3 + x) * (INTEGRAL e^(-x/3 - x) * x^3 dx + C)

This "(INTEGRAL e^(-x/3 - x)" is not integratable by any means I know of. So I e-mailed the Professor and this is what he told me to do.

"I gave this problem so you would have experience with something that doesn't have an explicit solution. The solution is in terms of an integral. You can even make it a definite integral with the upper limit variable. Choose the lower limit so the initial condition works out. Since your calculator can do numerical integration, you can then evaluate y(1)."

I am clueless on what he is talking about. I have looked in all of the DE texts I have and all say that it is fine to leave it in the form that it is in. However, none of those examples cover what to do when there are initial conditions present and you run into this problem.

Can anyone help me with this?

Thanks, Steve

Last edited:
How are you getting this step:

INT FACTOR = e^(INTEGRAL(-1-x^2) dx) = e^(-x/3 - x) so:

Using the integrating factor is given by the form:

dy/dx + A(x)y = B(x)

INT Factor = e^(A(x) dx)

so A(x) = (-1-x^2)

INt Factor is given by: e^(INTEGRAL (-1-x^2 dx)

INT Factor = e^(-x^3/3 - x)

Does that make sense?

Yes but e^(-x^3/3 - x) is different from e^(-x/3 - x) that you originally wrote. I guess I was hinting that you forgot the "^3" Does that make sense?

Fixed. Thanks.

Any idea how to the ODE?

Thanks, Steve

I wouldn't use integrating factor, probably method of Frobenius / power series method is what I'd jump to.

We haven't learned that yet. Any ideas as to what the Professor said?

Thanks for your help.

-Steve

## 1. What is an integrating factor in mathematics?

An integrating factor is a function that is used to solve differential equations by multiplying it with the given equation. This helps in simplifying the equation and making it easier to solve.

## 2. How do you find the integrating factor of a differential equation?

The integrating factor can be found by first identifying the type of differential equation and then using a specific formula or method to calculate it. For example, for first-order linear differential equations, the integrating factor is the exponential of the coefficient of the independent variable.

## 3. What is an unintegrable term in a differential equation?

An unintegrable term is a term in a differential equation that cannot be integrated using the standard integration techniques. These terms can make it difficult to solve the equation, and sometimes, numerical methods or approximations need to be used.

## 4. How do you handle unintegrable terms in a differential equation?

One way to handle unintegrable terms is to use numerical methods or approximations. Another way is to transform the equation using substitution or change of variables to eliminate the unintegrable term. In some cases, it may not be possible to solve the equation analytically and numerical solutions are needed.

## 5. Can all differential equations be solved using integrating factors?

No, not all differential equations can be solved using integrating factors. Integrating factors are usually used for first-order linear differential equations or equations that can be transformed into linear form. Other types of differential equations may require different methods or techniques for solving them.

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