Question 1046185
Applying implicit differentiation on {{{y^2=1+sin x}}}, we get

2yy' = cosx

another application of implicit differentiation yields

2[(y')^2 +yy"] = -sinx

===> (y')^2 +yy" = -(sinx)/2

===> (y')^2 +yy" + (sinx)/2 = 0, or

yy" + (y')^2 + (sinx)/2 = 0.

Since {{{y^2=1+sin x}}}, it follows that a = 1/2, and b = -1/2.