Use an integrating factor to solve

In summary, the equation is separable, but I don't know how to find P(x) and I don't know how to find a solution.
  • #1
Calu
73
0

Homework Statement



Use an integrating factor to determine the general solutions of the following differential
equation:

dx/dt - 2/t = 2t3 + (4t2)(e4t)


Homework Equations



R(x) = e∫P(x).dx

The Attempt at a Solution



Usually the equation is in the form dx/dt + P(x)t = Q(x) but here I'm not sure what to do to find P(x) here as I have 1/t, t3 and t2.

I'm also not sure how to go about finding a solution either. I know that once I have found the integrating factor, I have to multiply the original equation by R(x). After that I'm not sure what to do.
 
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  • #2
Well, it looks like this equation is actually separable. Just move the -2/t over, and the right side will be entirely a function of t.
 
  • #3
jackarms said:
Well, it looks like this equation is actually separable. Just move the -2/t over, and the right side will be entirely a function of t.

I realized that whilst I was doing it, but the question specifically asks me to use an integrating factor. thanks for the reply though.
 
  • #4
Calu said:

Homework Statement



Use an integrating factor to determine the general solutions of the following differential
equation:

dx/dt - 2/t = 2t3 + (4t2)(e4t)


Homework Equations



R(x) = e∫P(x).dx

The Attempt at a Solution



Usually the equation is in the form dx/dt + P(x)t = Q(x) but here I'm not sure what to do to find P(x) here as I have 1/t, t3 and t2.

I'm also not sure how to go about finding a solution either. I know that once I have found the integrating factor, I have to multiply the original equation by R(x). After that I'm not sure what to do.

Here, your independent variable is ##t##. So your integrating factor is ##R=e^{\int\frac 2 t~dt}##. Evaluate that and multiply through by it.
 
  • #5
LCKurtz said:
Here, your independent variable is ##t##. So your integrating factor is ##R=e^{\int\frac 2 t~dt}##. Evaluate that and multiply through by it.

Oh, I see. Thanks, I was being a bit stupid there.
 

Related to Use an integrating factor to solve

1. How do I determine which integrating factor to use?

The integrating factor for a differential equation is typically determined by examining the coefficients of the highest-order derivative and the variable being differentiated. If the coefficients are not constant or if there is a variable present, then an integrating factor is needed. The integrating factor is usually the coefficient of the variable being differentiated.

2. Can any differential equation be solved using an integrating factor?

No, not all differential equations can be solved using an integrating factor. The differential equation must be in a specific form, known as a "first-order linear" equation. This means that the highest-order derivative must have a coefficient that is a function of the independent variable and the variable being differentiated, and the remaining terms must be linear in the dependent variable.

3. What is the general process for using an integrating factor to solve a differential equation?

The general process for using an integrating factor to solve a differential equation is as follows:

  1. Write the differential equation in the form y' + P(x)y = Q(x)
  2. Determine the integrating factor, I(x), by taking the coefficient of y and raising it to the power of -1.
  3. Multiply both sides of the equation by I(x).
  4. Integrate both sides of the equation with respect to x.
  5. Solve for y by dividing both sides by I(x).

4. Are there any limitations to using an integrating factor to solve a differential equation?

Yes, there are a few limitations to using an integrating factor to solve a differential equation. One limitation is that the differential equation must be in a specific form, as mentioned in question 2. Additionally, the integrating factor may not always be easy to determine, and the process can become more complicated for higher-order differential equations. It is also important to note that integrating factors may not always provide an exact solution, but rather an approximate solution.

5. Can I use an integrating factor to solve a differential equation with initial conditions?

Yes, integrating factors can be used to solve differential equations with initial conditions. After solving for y using the integrating factor, the initial conditions can be used to find the specific values of any unknown constants that may arise in the solution. This allows for a more accurate and precise solution to the differential equation.

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