Losing a solution of a 1st order ODE?

  • #1
There is an example in a book regarding DEs which I do not understand. Solve the IVP

[tex] y'=y^2-4, y(0)=-2 [/tex]where t is the independent variable

We first solve by separation of variables to arrive at the 1-parameter solution.

[tex] -\frac{1}{4}ln (y+2)+\frac{1}{4}ln (y-2)=t+c[/tex]

Simplifying and expressing the solution explicitly, we find that,

[tex] y=2\frac{1+ce^{4t}}{1-ce^{4t}} [/tex]

Taking the initial condition,

[tex] -2=2\frac{1+c}{1-c} [/tex] which simplifies to,

-1=1.

They said that the solution is wrong because:

we can express the DE as, [tex] y'=(y+2)(y-2) [/tex] and that the when y=-2, and y=2 satisfies this equation (what does it mean?). How do we "preclude" y=-2 and y=-2 before solving starting to solve the DE?
 
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Answers and Replies

  • #2
nd
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Your initial condition y(0)=2 seems to imply that c = 0, not -1=1?
You can preclude y=-2 simply by noticing that it fails to satisfy the initial condition.
 
  • #3
sorry, my initial condition is [tex] y(0)=-2[/tex]. Do we ignore [tex]y=-2[/tex] because firstly, it doesn't tell us the value of c, and it gives us the illogical [tex]-1=1[/tex]?

Also, do we use the other alternative [tex]y=2[/tex] because it is the other "alternative" in [tex]y'=(y+2)(y-2)[/tex]??? If it is so, why?
 
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  • #4
arildno
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The "separable", non-constant solutions involve a division of both sides of the equation with the expression [tex]y^{2}-4[/tex]
But, since you can't divide by zero, you have implicitly assumed that [tex]y\neq\pm2[/tex]

In addition to the non-constant solutions, you've got the constant solutions [tex]y_{1}(t)=2,y_{2}(t)=-2[/tex]
 
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  • #5
dextercioby
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[tex] y(0)=-2 [/tex] is not a valid initial condition.When separating variables YOU ASUMED [tex] y\neq 2;y\neq -2 [/tex] Trying to impose the initial condition to

[tex] \frac{y-2}{y+2}=Ce^{4t} [/tex]

fails...

Daniel.
 

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