Question

given ivp y' = (2y)/x, y(x0) = y0

using the existence and uniqueness theorem show that a unique solution exists on any interval where x0 does not equal 0, no solution exists if y(0) = y0 does not equal 0, and and infinite number of solutions exist if y(0) = 0

Answer 1

qup da Solution- Given y's phone y (20) = Yo dy fra,y) = 2 By using existence ç uniqueness theorem, (a) Let f continuous region *# 0 continuous on region *#0 Ho to fy= 2 h is also an initial point (0o, you on region ako so, By using uniqueness theorem This initial value problem bag unique solution.

O on ty= 2 / (c) if not at not so Yo = cх² (b) Let Yoto f is continuouy region y is continuous on region yao & By using given Theorem This initial value problem baş no solution 20:40=0 f is continuous (0,0) also continuous This initial value problem does not have unique solution 2ye 2xcx? 2Cx : da and + 0 = you c1092 Ye = solutions for all Hence, This initial value problem has. infinety many solutions. Y = cх² dyc ane 4 * Thank you