I am given the following differential equation: $$x^2y'-y^2=1$$ where $y(1)=0$ and asked solve it:
lets divide the DE by $x^2$, $$y'-\frac{1}{x^2}\cdot y^2=\frac{1}{x^2}, x\ne0$$
Now lets find the find the integrating factor (IF):
$$e^{\int( -\frac{1}{x^2})dx}= e^{\frac{1}{x}}$$
Now lets multiply the DE by IF we get : $$(y^2\cdot e^{\frac{1}{x}})'=\frac{1}{x^2}\cdot e^{\frac{1}{x}}$$ hence $$y^2\cdot e^\frac{1}{x}= e^{\frac{1}{x}}+c$$ we know that $y(1)=0$ and we have already established that $x\ne 0$ we then get that $c=-1$ and hence the solution to the differential equation becomes: $y(x)= \pm \sqrt{1-\frac{1}{e^{\frac{1}{x}}}}$
Here is another problem though, i dont know when to choose the positive square root or the negative one for DEs. Is there any way to know what to choose?
Anyhow the books solution (i have the worked solution right in front of me here) says that: $$y(x)=tan(1-\frac{1}{x}), x>\frac{2}{2+\pi}$$
I fail to see what went wrong with my solution? and WHY is it wrong?
You write $(y^2 e^{\frac{1}{x}})'$ at one time, but notice that $$\frac{d}{dx} y^2 e^{\frac{1}{x}} = 2y y' e^{\frac{1}{x}} + y^2 e^{\frac{1}{x}} \frac{-1}{x^2}.$$ Do you see that $2y y'$ term there? That does not appear in your differential equation. In particular, you do not have a linear differential equation, so this technique won't work towards a solution.
It's worth mentioning that many of your questions would clear up by your checking the solution. Suppose that you did have two candidate solutions that looked like $\pm g(x)$. How can you determine which one (or if both) work? Plug them back into the differential equation and check!
This would reveal to you that your proposed solutions do not actually solve the differential equation as well.