How Do You Solve the Differential Equation dy/dx = 1 - y^2?

Classes/CalcII/PartialFractions.aspxIn summary, the conversation discussed the solution to a differential equation involving separation of variables and integration by partial fractions. The final solution for y was given as (x^2-4)/(x^2+4), with an initial condition applied to determine the constant of integration. The conversation also suggested reviewing integration by partial fractions for better understanding.
  • #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...
 
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  • #2
[tex]2\frac{dy}{1-y^2}=\frac{dy}{1-y}+\frac{dy}{1+y}[/tex]
 
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  • #3
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
BvU said:
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
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
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}##
 
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  • #7
chwala said:
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 !

##\ ##
 
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  • #8
BvU said:
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
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:

##\ ##
 
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  • #10
on integration i am getting,
chwala said:
##\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|}##

chwala said:
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 ...

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

1. What is a first order differential equation?

A first order differential equation is a mathematical equation that describes the relationship between a function and its derivative. It involves only the first derivative of the function and can be expressed in the form dy/dx = f(x).

2. How do you solve a first order differential equation?

To solve a first order differential equation, you can use various methods such as separation of variables, integrating factors, or the method of undetermined coefficients. It is important to first identify the type of equation and then choose the appropriate method for solving it.

3. What is the importance of solving first order differential equations?

First order differential equations are used to model various physical and natural phenomena in fields such as physics, engineering, and economics. Solving these equations allows us to understand and predict the behavior of these systems, making them an essential tool in scientific research and problem-solving.

4. Can first order differential equations have multiple solutions?

Yes, first order differential equations can have multiple solutions. This is because the general solution of a first order differential equation contains a constant of integration, which can take on different values and result in different solutions. However, for a specific initial condition, there will only be one unique solution.

5. What are some real-life applications of first order differential equations?

First order differential equations have numerous real-life applications, such as predicting population growth, modeling the spread of diseases, analyzing electrical circuits, and determining the rate of chemical reactions. They are also used in fields such as economics, biology, and environmental science to study various systems and phenomena.

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