Consider y′=y2−x2y' = y^2 - x^2y′=y2−x2. Given y1(x)=xy_1(x) = xy1(x)=x is not a solution, check if yp(x)=−xy_p(x) = -xyp(x)=−x is a solution.
Yes, because (−1)2−x2=−1≠−1(-1)^2 - x^2 = -1 \neq -1(−1)2−x2=−1=−1
No, because −1≠(−x)2−x2-1 \neq (-x)^2 - x^2−1=(−x)2−x2
Yes, because y′=−1y' = -1y′=−1 and y2−x2=(−x)2−x2=0y^2 - x^2 = (-x)^2 - x^2 = 0y2−x2=(−x)2−x2=0
No, because y′=−1y' = -1y′=−1 and y2−x2y^2 - x^2y2−x2 is not −1-1−1