Exercises 2.2Solving foryyieldsy=±1√1−x2.(2.6)Since, at the initial point,x= 0,y(0) = 1>1, we choose the positive sign in the aboveexpression fory. Thus, the solution isy=1√1−x2.Similarly we find solutions for the other two initial conditions:y(0) =12⇒C= 4⇒y=1√4−x2;y(0) = 2⇒C=14⇒y=1(1/4)−x2.(c)For the solution to the first initial problem in (b),y(0) = 1, the domain is the set of allvalues ofxsatisfying two conditions1−x2≥0(for existence of the square root)1−x2= 0(for existence of the quotient)⇒1−x2>0.Solving forx, we getx2<1⇒|x|<1or−1< x <1.In the same manner, we find domains for solutions to the other two initial value problems:
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