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

[murshid_islam]

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:
[tex]\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 [/tex]

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

So,
[tex]\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}[/tex]

Is this ok, or did I make any mistake?

 

[Orodruin]

Did you try inserting your solutions in the original ODE and boundary condition?

 

[murshid_islam]

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.

 

[murshid_islam]

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]

Yes.

 

[murshid_islam]

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]

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.

 

[murshid_islam]

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]

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

 

[murshid_islam]

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]

But my first post mentions only y=x and y=1/x

I understand , but notice your solution included a ## +/- ##

 

[murshid_islam]

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]

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

 

[murshid_islam]

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]

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?

My bad, i got confused by the +/- sign. Please disregard.