Differentials with linear equations

  • Thread starter maiad
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  • #1
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Homework Statement


Suppose that y(x) is the solution to the initial problem, y'=y(1-x), y(1)=e
find y(2)


Homework Equations





The Attempt at a Solution


This is my initial attempt:
[itex]\frac{dy}{dx}[/itex]=y(1-x)
[itex]\frac{dy}{y}[/itex]=(1-x)dx

i then integrated both side to get:
lny=-ln(1-x)+C

and here's the problem, if i plug in 1 to find c, ln(1-x) does not exist at that point....
 
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Answers and Replies

  • #2
102
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I tried another approach:
y'+yx=y
(yx)'=y

integrated both sides:
yx=[itex]\frac{1}{2}[/itex]([itex]y^{2}[/itex]) +C

but with this approach i wasn't able to solve for y
 
  • #3
1,552
15
[itex]\frac{dy}{y}[/itex]=(1-x)dx

i then integrated both side to get:
lny=-ln(1-x)+C
First of all, you integrated the right hand side incorrectly. Remember, you're just integrating a polynomial.
 
  • #4
1,552
15
I tried another approach:
y'+yx=y
(yx)'=y
Wait a minute, (yx)'=xy'+yx'=xy'+y, not y'+yx.
 

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