Relax

Respuesta :

Answer :

2y=1+e−x2

Solution :

IF=e∫2xdx=ex2

∴y×ex2=∫xex2dx=12∫etdt+C, where x2=t

=12ex2+C.

∴y=12+Cex2

Putting y=1andx=0, we get C=12.