Let y=y(x) solve dxdy=(y+1)[(y+1)ex2/2−x], 0<x<2, y(2)=0. Then y′(1) is:
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Official solution
With t=y+11, a Bernoulli substitution gives dxdt−xt=−ex2/2, whose I.F. is e−x2/2. Solving and using y(2)=0 (i.e. t(2)=1) yields, after differentiating, y′(1)=−(e2+1)2e3/2.