Q.
If the general solution of the differential equation y′=xy+Φ(yx), for some function Φ, is given by yln∣cx∣=x, where c is an arbitrary constant, then Φ(2) is equal to :
y′=xy+ϕ(yx)...(1) is solution of yIn∣cx∣=x...(2)
d.w.r. to x ∣cx∣∣y⋅cx∣∣cx∣⋅c+In∣cx∣y′=1 xy+yxy′=1(useInc∣cx∣=yx) y′=(1−xy)xy
use y' in equation (1) xy(1−xy)=xy+ϕ(yx)
put (yx)=2⇒(1−21)(21)=21+ϕ(2) =41=21+ϕ(2) ϕ(2)=−41