Separable Differential Equation Question

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SUMMARY

The discussion centers on the separable differential equation dy/dx = (y^2 - 1)/(x^2 - 1) with the initial condition y(2) = 2. Participants clarify that while y = 1 and y = -1 are solutions to the differential equation, they do not satisfy the initial condition, thus excluding them as valid solutions for the problem at hand. The separation of variables method is employed, leading to the conclusion that these values must be excluded due to their failure to meet the specified boundary condition.

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inter060708
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Homework Statement


dy/dx = (y^2 - 1)/ (x^2 - 1) with initial condition y(2) = 2

Why is y = 1 and/or y= -1 not solutions?


Homework Equations





The Attempt at a Solution



I am actually able to solve this differential equation but when I separate the equation according to x and y:

(y^2 - 1)^-1 dy = (x^2 - 1)^-1 dx

here I am dividing both sides of the equation by (y^2 - 1), which means now I have to exclude y = 1 , -1 and check whether they are solutions

Now how do I show that y =1 and y = -1 are not solutions?
 
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inter060708 said:

Homework Statement


dy/dx = (y^2 - 1)/ (x^2 - 1) with initial condition y(2) = 2

Why is y = 1 and/or y= -1 not solutions?

Homework Equations


The Attempt at a Solution



I am actually able to solve this differential equation but when I separate the equation according to x and y:

(y^2 - 1)^-1 dy = (x^2 - 1)^-1 dx

here I am dividing both sides of the equation by (y^2 - 1), which means now I have to exclude y = 1 , -1 and check whether they are solutions

Now how do I show that y =1 and y = -1 are not solutions?

y=1 or y=(-1) don't satisfy y(2)=2, do they? They are solutions to the differential equation but they don't satisfy your boundary condition.
 

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