Integrating factor problems (property of exponents)

  • Thread starter darryw
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  • #1
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Homework Statement



my integrating factor for the DE ty' + (t+1)y = t is mu(x) = e^integ (1+(1/t))
= e^(t + ln|t| + c)
so does this simplify to this...or not?

= e^t + t + c

so that DE becomes:

((e^t) + t))y = (e^t) + t)

and then after integrating...

((e^t) + t))y = e^t + (1/2)t^2 + c

is that correct??

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
35,017
6,767

Homework Statement



my integrating factor for the DE ty' + (t+1)y = t is mu(x) = e^integ (1+(1/t))
= e^(t + ln|t| + c)
For the integrating factor you don't need the constant.
so does this simplify to this...or not?

= e^t + t + c
Or not. Your integrating factor is et + ln(t). This is NOT EQUAL to et + eln(t). There's a property of exponents you need here that can be used to simplify ab + c. What is it?
so that DE becomes:

((e^t) + t))y = (e^t) + t)

and then after integrating...

((e^t) + t))y = e^t + (1/2)t^2 + c

is that correct??

Homework Equations





The Attempt at a Solution

 
  • #3
127
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ty' + (t+1)y = t initial conditions: y(ln(2)) = 1

mu(x) = e^integ (1+(1/t))
= e^(t + ln|t|)
= te^t

(te^t)y = (te^t)t

integ ((te^t)y)' = integ(t^2)(e^t)

(te^t)y = e^t(t^2 - 2t +2 + c)

ty = t^2 - 2t +2 + c

y = t-2 + (2/t) + c/t
 
  • #4
35,017
6,767
ty' + (t+1)y = t initial conditions: y(ln(2)) = 1
Your next step was to divide both sides by t to get
y' + (1 + 1/t)y = 1 (*)

This is the equation you used to get your integration factor.
mu(x) = e^integ (1+(1/t))
= e^(t + ln|t|)
= te^t

(te^t)y = (te^t)t
Too many t factors on right side. It should be just te^t.

The whole reason for find the integration factor is to be able to recognize the left side as the derivative of something. Multiplying (*) on both sides by te^t, we get
y' te^t + (1 + 1/t)y te^t = te^t

or
y' te^t + (y te^t + y e^t)= te^t

The left side happens to be the derivative of y*te^t, which you can check using the product rule.

So what we have is
d/dt[y te^t] = te^t
[tex]\Rightarrow y te^t = \int t e^t dt[/tex]

Can you continue from here?
integ ((te^t)y)' = integ(t^2)(e^t)

(te^t)y = e^t(t^2 - 2t +2 + c)

ty = t^2 - 2t +2 + c

y = t-2 + (2/t) + c/t
 
  • #5
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I want to clarify why i need to use an integrating factor, to see if im understanding...
This equation is not separable, that is i cant separate all the t's from the y's. If it were separable it is solvable. correct?
ty' + (t+1)y = t
It is not exact either, so the point of using integrating factor is to make the equation exact. And once exact, I can then follow the steps for solving exact equation. correct?

To be exact, then M_x(x,y) + N_y(x,y) = 0.. That is, derivative of function wrt x + derivative of function wrt y = 0.

is this correct?
 

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