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

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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Orodruin
Staff Emeritus
Homework Helper
Gold Member
Did you try inserting your solutions in the original ODE and boundary condition?

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.

Michael Price
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?

fresh_42
Mentor
Yes.

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?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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.

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?

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

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

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

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} ##

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

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?

WWGD