Solve this first order differential equation

  • Thread starter chwala
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  • #1
chwala
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Homework Statement:
See attached
Relevant Equations:
understanding of integration and separation of variables.
This is the question;

1640349320134.png



This is the solution;
1640349379745.png


Find my approach here,

##x####\frac {dy}{dx}##=##1-y^2##
→##\frac {dx}{x}##=##\frac {dy}{1-y^2}##
I let ##u=1-y^2## → ##du=-2ydy##, therefore;
##\int ####\frac {dx}{x}##=##\int ####\frac {du}{-2yu}##, we know that ##y##=##\sqrt {1-u}##
##\int ####\frac {dx}{x}##=##\int ####\frac {du}{-2u\sqrt {1-u}}## i let,
##\frac {1}{u\sqrt {1-u}}##=##\frac {A}{\sqrt {1-u}}##+##\frac {B}{u}##
→##1=##Au##+##B##\sqrt {1-u}##
##A=0.5## and ##B=0.5## * i need to check how to arrive at this...i got a bit stuck here...
Therefore,
##\frac {1}{-2}##[##\int####\frac {0.5}{\sqrt {1-u}}####du##+##\int####\frac {0.5}{u}]####du##=##\int ####\frac {dx}{x}##
##\frac {1}{-4}####\int####\frac {1}{\sqrt {1-u}}####du##+##\frac {1}{-4}####\int####\frac {1}{u}####du##=##\int ####\frac {dx}{x}##
on integration we shall have,
##-0.25(1-(1-y^2))-0.25 ln|1-y^2|##=##ln|x|## + ##k##
##-0.25y^2-0.25ln|1-y^2|##=##ln|x|## + ##k##
using and applying the initial conditions ##y(2)=0##, we get,
##k=-ln2##

i will need to re check this later...something does not look right...i will amend this post to correct solution then look at the suggested approach...
 
Last edited:

Answers and Replies

  • #2
anuttarasammyak
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[tex]2\frac{dy}{1-y^2}=\frac{dy}{1-y}+\frac{dy}{1+y}[/tex]
 
  • #3
BvU
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Hello @chwala !

Your marking scheme wants separation of variables. I empathically suggest you try that: all ##y ## stuff on the left and all ##x## stuff on the right !

Then @anuttarasammyak's giveaway should help you integrate both sides.

##\ ##
 
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  • #4
chwala
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Hello @chwala !

Your marking scheme wants separation of variables. I empathically suggest you try that: all ##y ## stuff on the left and all ##x## stuff on the right !

Then @anuttarasammyak's giveaway should help you integrate both sides.
I suppose he/she means $$\frac{1}{1-y^2}=\frac{1}{1-y}+\frac{1}{1+y}\quad ?$$

##\ ##
True, arrrrgh my brain went for a walk:biggrin::biggrin:...difference of two squares right is much faster,i will nevertheless try and complete on what i had started. My presumption is that we should get same solution.
 
  • #5
chwala
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I tend to think my approach was wrong,...it would be difficult to ascertain the values of the constants after having decomposed into the partial fractions...the unknowns in my working are ##3## variables, ##A,B## and ##u##...
Thanks bvu and anuttarasa... with the given suggested approach, then the working to solution would be easy. Cheers guys
 
  • #6
chwala
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Ok from this step, i have

##x####\frac {dy}{dx}##=##1-y^2##

→##\frac {dx}{x}##=##\frac {dy}{1-y^2}##

##\int ####\frac {dx}{x}##=##\int ####\frac {dy}{({1+y})({1-y})}##

Let, ##\frac {1}{({1+y})({1-y})}##=##\frac {A}{1-y}##+##\frac {B}{1+y}##

##1=A(1+y)+B(1-y)##

→We get the simultaneous equation,
##A+B=1##
##A-B=0##

##A=0.5## and ##B=0.5## therefore we shall have

##\int ####\frac {dx}{x}##=##\int ####\frac {0.5dy}{1-y}##+##\int ####\frac {0.5dy}{1+y}##

...on integration we get,

##\frac {1}{2}####ln|1+y|####-\frac {1}{2}####ln|1-y|##=##ln|x|##+##k##

on applying ##y(2)=0##,

we get ##k##=##-ln|2|## therefore,

##\frac {1}{2}####ln|1+y|####-\frac {1}{2}####ln|1-y|##=##ln|x|####-ln|2|##

##\frac {1}{2}####ln####\frac {|1+y|}{|1-y|}##=##ln####\frac {|x|}{|2|}##→

→...##\frac {1+y}{1-y}##=##\frac {x^2}{4}##

##4y+x^2y=x^2-4##

##y(4+x^2)=x^2-4##

##y##=##\frac {x^2-4}{x^2+4}##
 
Last edited:
  • #7
BvU
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Ok from this step, i have
##x####\frac {dy}{dx}##=##1-y^2##
→##\frac {dx}{x}##=##\frac {dy}{1-y^2}##
##\int ####\frac {dx}{x}##=##\int ####\frac {du}{1-y}####⋅####\frac {dy}{1+y}## ?:)?:)?:)
Let, ##\frac {1}{({1+y})({1-y})}##=

About to run off the rails ... :nb)
First: see the repaired #2 for a missing 2.
Second: the integral of a sum is a sum of integrals, not a product !

##\ ##
 
  • #8
chwala
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About to run off the rails ... :nb)
First: see the repaired #2 for a missing 2.
Second: the integral of a sum is a sum of integrals, not a product !

##\ ##
still posting...learning latex slowly...thanks bvu..merry xmas.
 
  • #9
BvU
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Merry xmas to you too ! Actually, we all should have better things to do at this moment, but PF is a bit of an addiction :biggrin:

##\ ##
 
  • #10
BvU
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on integration i am getting,
##\frac {1}{2}####ln|1+y|##-##\frac {1}{2}####ln|1-y|##=##ln|x|##+##k##
on applying ##y(2)=0##,
we get ##k##=##-ln|2|## therefore,
##\frac {1}{2}####ln|1+y|##-##\frac {1}{2}####ln|1-y|##=##ln|x|##+##-ln|2|##
##\frac {1}{2}####ln####\frac {|1+y|}{|1-y|}##=##ln####\frac {|x|}{|2|}##

learning latex slowly...
It's the only way !
Tips (unasked for :rolleyes:, but I can't help it :wink: ):
  • omit all the quadruple #### -- it makes it easier to read for you (and to copy for others :biggrin: ).
  • Once expressions get a little bigger, displayed math (enclosed in $$) improves readability
and, sure enough, once things are put back on the rails, your proceed well, up to

$$ {{1+y}\over {1-y}} = {x^2\over 4}$$ where a new derailing occurs ...

##\ ##
 
  • #11
BvU
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The result is now impeccable ! Scrap the last part of #10 !
$$y=\frac {x^2-4}{x^x+4}$$:partytime:Bingo !:partytime:
 
  • #12
chwala
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Bingo!Merry xmas guys...Going for a cup of coffee at the restaurant:cool::cool::cool:
 
  • #13
MidgetDwarf
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Review integration by partial fractions for the integral involving y.
 

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