# Differential equation problem: Solve dy/dx = (y^2 - 1)/(x^2 - 1), y(2) = 2

• #1
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Homework Statement:
Solve dy/dx = (y^2 - 1)/(x^2 - 1), y(2) = 2
Relevant Equations:
dy/dx = (y^2 - 1)/(x^2 - 1), y(2) = 2
This is my attempt:
$$\frac{dy}{dx} = \frac{y^2 - 1}{x^2 - 1} \\ \int \frac{dy}{y^2 - 1} = \int \frac{dx}{x^2 - 1} \\ \ln \left| \frac{y-1}{y+1} \right| + C_1 = \ln \left| \frac{x-1}{x+1} \right| + C_2 \\ \ln \left| \frac{y-1}{y+1} \right| = \ln \left| \frac{x-1}{x+1} \right| + C$$

Since y(2) = 2,
$$\ln \left| \frac{1}{3} \right| = \ln \left| \frac{1}{3} \right| + C \\ \therefore C = 0$$

So,
$$\ln \left| \frac{y-1}{y+1} \right| = \ln \left| \frac{x-1}{x+1} \right| \\ \left| \frac{y-1}{y+1} \right| = \left| \frac{x-1}{x+1} \right| \\ \frac{y-1}{y+1} = \pm \frac{x-1}{x+1} \\y = x, y = \frac{1}{x}$$

Is this ok, or did I make any mistake?

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• #2
Orodruin
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Did you try inserting your solutions in the original ODE and boundary condition?

• #3
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Did you try inserting your solutions in the original ODE and boundary condition?
Both y = x and y = 1/x satisfy the original ODE, but the boundary condition works for y = x only.

Likes Michael Price
• #4
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Both y = x and y = 1/x satisfy the original ODE, but the boundary condition works for y = x only.
Is that it? Is that why y = x is the only solution?

• #5
fresh_42
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Yes.

• #6
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Both y = x and y = 1/x satisfy the original ODE, but the boundary condition works for y = x only.
Is that it? Is that why y = x is the only solution?
If the boundary condition was ##y(1) = 1##, both ## y = x ## and ## y = \frac{1}{x} ## would be correct solutions, right?

• #7
Orodruin
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If the boundary condition was ##y(1) = 1##, both ## y = x ## and ## y = \frac{1}{x} ## would be correct solutions, right?
The differential equation is not well defined in (x,y) = (1,1) as you have an expression of the form 0/0 for dy/dx.

• #8
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The differential equation is not well defined in (x,y) = (1,1) as you have an expression of the form 0/0 for dy/dx.
Oh yes. How did I not notice this?

• #9
WWGD
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Murshid, did you check the other possibility of y=-x (from your first post) ?

• #10
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Murshid, did you check the other possibility of y=-x (from your first post) ?
But my first post mentions only y=x and y=1/x

• #11
WWGD
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But my first post mentions only y=x and y=1/x
I understand , but notice your solution included a ## +/- ##

• #12
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I understand , but notice your solution included a ## +/- ##
Yes, ## \frac{y-1}{y+1} = \frac{x-1}{x+1} ## leads to ## y = x ##, and ## \frac{y-1}{y+1} = - \frac{x-1}{x+1} ## leads to ## y = \frac{1}{x} ##

• #13
WWGD
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Ok, it is not a solution to the original, I just wanted to ask if you had tested it.

• #14
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Ok, it is not a solution to the original, I just wanted to ask if you had tested it.
Should I? I don't get this. Why should I test it if it's not a solution to the differential equation? Are you hinting something that I'm missing?

• #15
WWGD
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Should I? I don't get this. Why should I test it if it's not a solution to the differential equation? Are you hinting something that I'm missing?

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