Differential Equation using Integrating Factor

In summary, the ODE given is (1+x)*dy/dx - xy = x+x^2 and the method of solution is using the integration factor. The first step is to divide both sides by (1+x) and then factor out the x on the LHS. Then, dividing both sides by x and moving the -y/(1+x) to the RHS produces dy/dx = y/(1+x). The next step is to use the product rule to find the integrating factor, which is u(x) = 1+x. After integrating the separable equation, the integrating factor becomes 1+u. Finally, upon integration, the solution is y = c/(1+x).
  • #1
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Homework Statement



The ODE is

(1+x)*dy/dx - xy = x+x^2

Homework Equations



The method of solution is to be through the use of the integration factor.

The Attempt at a Solution



First, I divided each side by (1+x) to produce

dy/dx - xy/(1+x) = x

then factor out the x on the LHS to produce

dy/dx - x * (y/(1+x)) = x

then divide both sides by x to produce

dy/dx - y/(1+x) = 0

now move the -y/(1+x) to the RHS to produce

dy/dx = y/(1+x)

That is all the farther I have progressed. Am I correct so far?

Thanks
Matt
 
Last edited:
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  • #2
Yes, you are correct so far by what progress is that? How does that help you find the integrating factor?

The left side of your equation is (1+x)(dy/dx)- xy. The whole point of an "integrating factor" is that it is a function u(x) such that u(x)(1+x)(dy/dx)- uxy= d(u(x)(1+x)y)/dx.
Go ahead and use the product rule on the right side. That gives u(1+x)(dy/dx)+ u(1)y+ (du/dx)(1+x)y and that must be equal to u(x)(1+x)(dy/dx)- uxy. Well, the "u(1+x)(dy/dx)" cancels immediately leaving u(x)y+ (1+x)y(du/dx)= -uxy so (1+x)y(du/dx)= -xy(1+ u). Now the "y"s cancel so (1+x)(du/dx)= -x(1+u) which is a separable equation. du/(1+u)= -dx/(x(1+x)). Integrate that to find the integrating factor.
 
  • #3
By progress, I was meaning the reformatting of the original equation. I believed that this equation is seperable but the assignment was to use the integrating factor method.

I would rather take the path of the seperable equation.

Starting with

dy/dx = y/(1+x)

multiplying both sides by dx produces

dy = y/(1+x)dx

dividing both sides by y produces

dy/y = dx/(1+x)

integrating both sides produces

ln(y) = ln(1+x) + c

Is this correct so far?


Now proceeding forward with your instructions to integrate

du/(1+u) = -dx/(x(1+x)

upon integration of both sides

ln(1+u) = -ln(x)+ln(1+x)

now the integrating factor would be

e^(ln(1+u))

upon integration would be

1+u

Am I correct so far?

Thanks
Matt
 
Last edited:
  • #4
OK. I think I have the solution.

With the integrating factor of 1+x and using the equation

dy/dx -y/(1+x) = 0

multiplying both sides by the integrating factor produces

(1+x)*dy/dx - (1+x)*y/(1+x) = 0

which is equal to

(1+x) * dy/dx - y = 0

upon integrating

(1+x)*y = c

solving for y produces

y = c/(1+x)

Am I correct?

Thanks
Matt
 

1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a quantity and its rate of change. It contains one or more derivatives of the unknown quantity and is typically used to model real-world phenomena in physics, engineering, and other fields.

2. What is an integrating factor?

An integrating factor is a function used to solve a differential equation by transforming it into an integrable form. It is typically used for first-order linear differential equations, where the integrating factor is multiplied to both sides of the equation to make it easier to integrate.

3. How do you find the integrating factor for a differential equation?

The integrating factor for a first-order linear differential equation can be found by taking the exponential of the integral of the coefficient of the unknown variable. If the equation is not in standard form, it may need to be manipulated before finding the integrating factor.

4. What is the purpose of using an integrating factor?

The purpose of using an integrating factor is to transform a differential equation into an integrable form. This allows for easier solution and can also help to find a general solution that satisfies all possible initial conditions.

5. Can an integrating factor be used for all types of differential equations?

No, an integrating factor is typically used for first-order linear differential equations. Other types of differential equations, such as second-order or non-linear equations, may require different methods for solving.

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