Solving Problem 1: 3x' + 1/t x = t

  • Thread starter ollyfinn
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In summary, when integrating a function 1/t times a term with an exponent 1, you must use the ln function to simplify the equation.
  • #1
ollyfinn
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I have the following two questions to solve

Problem 1.

3x' + 1/t x = t

and

Problem 2.

x' + 1/t x = ln t

I have followed a method detailed in my textbook to try and get an answer for Problem 1 but am a bit unsure so if anyone can clarify my workings below before I spend time trying to solve Problem 2.

3x' + 1/t x = t

Fits the format dx/dt + g(t)x = f(t)

For the integrating factor I(t) = e^∫g(t) dt

∫1/t = ln t

e^ln t = t

Multiply both sides by I(t) so the equation becomes d/dt(I(t)x(t)) = I(t)f(t)

3 d/dt (tx) = t^2

Then I(t)x(t) = ∫I(t)f(t) dt + C

3tx = ∫t^2 dt + C

3tx = (t^3)/3 + C

x(t) = ((t^3)/3 + C)/3t

x(t) = 1/6t^2 + C1/3t^-1

Does this look right? If so I will attempt Problem 2. Thanks for any assistance.
 
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  • #2
ollyfinn said:
I have the following two questions to solve

Problem 1.

3x' + 1/t x = t

and

Problem 2.

x' + 1/t x = ln t

I have followed a method detailed in my textbook to try and get an answer for Problem 1 but am a bit unsure so if anyone can clarify my workings below before I spend time trying to solve Problem 2.

3x' + 1/t x = t

Fits the format dx/dt + g(t)x = f(t)

For the integrating factor I(t) = e^∫g(t) dt

∫1/t = ln t

e^ln t = t

Multiply both sides by I(t) so the equation becomes d/dt(I(t)x(t)) = I(t)f(t)

3 d/dt (tx) = t^2

Then I(t)x(t) = ∫I(t)f(t) dt + C

3tx = ∫t^2 dt + C

3tx = (t^3)/3 + C

x(t) = ((t^3)/3 + C)/3t

x(t) = 1/6t^2 + C1/3t^-1
3*3 is NOT 6! Other than that, and the need for parentheses to make it clearer (most people would interpret "1/6t^2" as "1/(6t^2)" and I don't think that is what you mean), you are doing this correctly.

Does this look right? If so I will attempt Problem 2. Thanks for any assistance.
 
  • #3
You are right my parentheses does need to be clearer.

My final answer should be:

x(t) = 1/9(t^2) + C1/3(t^-1)

Does that look better?
 

1. What is the first step in solving problem 1?

The first step in solving problem 1 is to simplify the equation by combining like terms. This means combining any constants and variables on the same side of the equation.

2. How do I isolate the variable in problem 1?

To isolate the variable, you must get it by itself on one side of the equation. This can be done by using inverse operations, such as adding or subtracting, to move other terms to the other side of the equation.

3. What is the next step after isolating the variable?

Once the variable is isolated, you can solve for it by using the appropriate operations to get it to equal a specific value. For example, in this problem, you may need to use the distributive property to get rid of any parentheses before solving for the variable.

4. Is there more than one way to solve problem 1?

Yes, there are multiple ways to solve problem 1. One method is to use algebraic techniques, as described above. Another method is to graph the equation and find the point(s) of intersection. You can also use a calculator to solve the equation numerically.

5. How do I know if my solution is correct?

To check if your solution is correct, you can substitute the value of the variable back into the original equation and see if it satisfies the equation. You can also graph the equation and see if the point of intersection matches your solution. Additionally, you can use a calculator to verify your solution numerically.

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