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:
\frac{dy}{dx} = \frac{y^2 - 1}{x^2 - 1}<br /> \\ \int \frac{dy}{y^2 - 1} = \int \frac{dx}{x^2 - 1}<br /> \\ \ln \left| \frac{y-1}{y+1} \right| + C_1 = \ln \left| \frac{x-1}{x+1} \right| + C_2<br /> \\ \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<br /> \\ \therefore C = 0

So,
\ln \left| \frac{y-1}{y+1} \right| = \ln \left| \frac{x-1}{x+1} \right|<br /> \\ \left| \frac{y-1}{y+1} \right| = \left| \frac{x-1}{x+1} \right|<br /> \\ \frac{y-1}{y+1} = \pm \frac{x-1}{x+1}<br /> \\y = x, y = \frac{1}{x}

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.