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If ( yx )= e|x|+ 2 is the solution of the differential equation x ( yx ) dydx=y ( yx )+x and y(1)= 3, then 2 is equal to
(4) 3
( yx )(x dy-y dx)=x dx; divide by x2:
( yx ) x dy-y dxx2= dxx
∵ x dy-y dxx2=d ( yx ), put yx=t: t dt= dxx
( yx )= e|x|+c, so c= 2
y(1)= 3: 3=0+ 2 = 3
2=3
Answer: (4)
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