Is this solution correct? I thought it was meant to be $y = 1/t-c_i,$ if the solution is wrong, what's the correct answer? The question is solve the following ODE: $t^2y'' =(y')^2 ,\; t>0.$
Here's the solution file:///M:/pc/My%20Documents/Doc3.pdf 
This is the solution my lecture provided and I think there's an error somewhere when $c_1$ doesn't equal $0;$ shouldn't it be $y= t^2/2+\ln (1-c_1t) + c_2?$
If the solution is wrong, what's the correct solution? Thanks: spent over 6 hours trying to work it out to come to the conclusion that there's an error. 

Hint: Substitute $$y''=v'$$ then you will get $$\frac{t^2v'(x)}{v(x)}=1$$
