D.E. Linear equation with integrating factor

In summary: I'm glad it worked!In summary, the correct solution for the given differential equation is y(t) = 7tet + 8e8t - 6et and the software may require the use of exp(t) notation instead of the standard et.
  • #1
Marco Lugo
9
0

Homework Statement



https://webwork.utpa.edu/webwork2_files/tmp/equations/2d/02a7e6a06f5b2424758fa01cc965f71.png with https://webwork.utpa.edu/webwork2_files/tmp/equations/80/81c176aa8964438a63eb096513245f1.png

Homework Equations


[/B]
Standard form: y' + p(x)y = f(x)

Integrating factor: r(x)= e∫p(x)dx

Solution: y(x) = e-∫p(x)dx(∫e∫p(x)dx ⋅ f(x) dx + C)

The Attempt at a Solution



From, y' - y = 7et + 56e8t

I got, p(x) = -1 and f(x) = 7et + 56e8t

So the integrating factor is r(t)= e∫-1 dt = e-t

Plug it into the solution formula,
y(t)= et [∫e-t (7et + 56e8t) dt + C]

= et (7t + 8e7t+ C)

= 7tet + 8e8t+ Cet

Using the initial condition y(0)= 2

= 8 + C = 2
C = -6

My final solution,
y(t) = 7tet + 8e8t- 6et

check: y(0) = 7(0)e(0) + 8e8(0)- 6e(0) = 2

But when I type it into WeBWork, it says my answer is incorrect. I don't think its a syntax problem because it has a preview button that allows you to see your actual problem.

Could it be that, y' - y = 7et + 56e8t is not in standard form?
 
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  • #2
Marco Lugo said:

Homework Statement



https://webwork.utpa.edu/webwork2_files/tmp/equations/2d/02a7e6a06f5b2424758fa01cc965f71.png with https://webwork.utpa.edu/webwork2_files/tmp/equations/80/81c176aa8964438a63eb096513245f1.png

Homework Equations


[/B]
Standard form: y' + p(x)y = f(x)

Integrating factor: r(x)= e∫p(x)dx

Solution: y(x) = e-∫p(x)dx(∫e∫p(x)dx ⋅ f(x) dx + C)

The Attempt at a Solution



From, y' - y = 7et + 56e8t

I got, p(x) = -1 and f(x) = 7et + 56e8t

So the integrating factor is r(t)= e∫-1 dt = e-t

Plug it into the solution formula,
y(t)= et [∫e-t (7et + 56e8t) dt + C]

= et (7t + 8e7t+ C)

= 7tet + 8e8t+ Cet

Using the initial condition y(0)= 2

= 8 + C = 2
C = -6

My final solution,
y(t) = 7tet + 8e8t- 6et

check: y(0) = 7(0)e(0) + 8e8(0)- 6e(0) = 2

But when I type it into WeBWork, it says my answer is incorrect. I don't think its a syntax problem because it has a preview button that allows you to see your actual problem.

Could it be that, y' - y = 7et + 56e8t is not in standard form?
I don't see anything wrong with your work or your solution. You can (and should) verify for yourself that your solution is correct by first checking that the initial condition is satisfied, and then by substituting your solution into the differential equation. Your solution should make the DE a true statement.

As far as why the software doesn't accept your solution, make sure that you are working the same problem it thinks you are working. I would also check with the instructor.
 
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  • #3
Mark44 said:
I don't see anything wrong with your work or your solution. You can (and should) verify for yourself that your solution is correct by first checking that the initial condition is satisfied, and then by substituting your solution into the differential equation. Your solution should make the DE a true statement.

As far as why the software doesn't accept your solution, make sure that you are working the same problem it thinks you are working. I would also check with the instructor.

I checked the initial condition but you're right I should have substituted my solution into the problem.

my solution,
y(t) = 7tet + 8e8t - 6et

check:
dy/dt= 64e8t + 7tet + et

substituting both into the initial probelm,

dy/dt - y = 56e8t + 7et

64e8t + 7tet + et - 7tet - 8e8t + 6et

I get, 56e8t + 7et

I don't see anything wrong. I would submit the answer again but I only have 2 more tries, so I want to get some more advice before I do.
 
  • #4
Marco Lugo said:
I checked the initial condition but you're right I should have substituted my solution into the problem.

my solution,
y(t) = 7tet + 8e8t - 6et

check:
dy/dt= 64e8t + 7tet + et

substituting both into the initial probelm,

dy/dt - y = 56e8t + 7et

64e8t + 7tet + et - 7tet - 8e8t + 6et

I get, 56e8t + 7et

I don't see anything wrong. I would submit the answer again but I only have 2 more tries, so I want to get some more advice before I do.

Your solution is correct.

Is there some special format you must adhere to when submitting solutions? For example, do you need to say exp(t) instead of et? Does it matter whether the terms come in a particular order, different from the one you wrote? For example, would ##8 e^{8t} + 7t e^{t} - 6e^{t}## or ##8 e^{8t} - 6 e^{t} + 7 t e^{t}## both be acceptable? Can you write ##t e^{t}## (or ##t \exp(t)##), or must it be in the form ##e^{t} \, t## (or ##\exp(t)\, t##), etc? I would certainly hope that the software would not be so picky, but who knows?
 
  • #5
Ray Vickson said:
Your solution is correct.

Is there some special format you must adhere to when submitting solutions? For example, do you need to say exp(t) instead of et? Does it matter whether the terms come in a particular order, different from the one you wrote? For example, would ##8 e^{8t} + 7t e^{t} - 6e^{t}## or ##8 e^{8t} - 6 e^{t} + 7 t e^{t}## both be acceptable? Can you write ##t e^{t}## (or ##t \exp(t)##), or must it be in the form ##e^{t} \, t## (or ##\exp(t)\, t##), etc? I would certainly hope that the software would not be so picky, but who knows?

It worked! I had to use the exp(t) notation; which is weird because the software usually isn't picky and it would say if it wanted something expressed in a certain way.
Anyway thanks Mark44 and Ray Vickson for all you're help :D
 

1. What is a D.E. linear equation with integrating factor?

A D.E. linear equation with integrating factor is a type of first-order differential equation that involves a linear combination of the dependent variable and its derivative, along with an integrating factor that is used to simplify the equation and make it easier to solve.

2. How is an integrating factor determined in a D.E. linear equation?

The integrating factor is determined by multiplying the entire equation by a suitable function that makes the left side of the equation equal to the derivative of the product of the integrating factor and the dependent variable. This function is typically determined by using the method of integrating factors, which involves finding the solution to an auxiliary equation.

3. What is the purpose of an integrating factor in a D.E. linear equation?

The purpose of an integrating factor is to simplify the equation and make it easier to solve. By multiplying the equation with the integrating factor, the left side of the equation can be transformed into the derivative of the product of the integrating factor and the dependent variable, which can then be integrated to find the solution to the original equation.

4. Are all D.E. linear equations solvable using an integrating factor?

No, not all D.E. linear equations are solvable using an integrating factor. The equation must be in a specific form, with a linear combination of the dependent variable and its derivative, in order for the method of integrating factors to be applicable. Additionally, some equations may require other methods or techniques to solve.

5. What are the steps for solving a D.E. linear equation with an integrating factor?

The steps for solving a D.E. linear equation with an integrating factor are as follows: 1) Identify the equation as a first-order linear differential equation; 2) Determine the integrating factor by multiplying the equation with a suitable function; 3) Rewrite the equation in terms of the integrating factor and the dependent variable; 4) Integrate both sides of the equation; 5) Solve for the dependent variable to find the general solution; 6) If given initial conditions, use them to find the particular solution.

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