GIA'I PHU.O.NG TRNH

6)

Gia'i phu.o.ng trnh:

^{y”(1 +y) =y}

⁰²

^+y

⁰

HD gia’i:

D - a.t

^y

⁰

^=z(y)⇒z

⁰

^=zdz_dy

thay vao phu.o.ng trnh:

^dzz+ 1 = dyy+ 1⇒z+ 1 =C

1

(y+ 1) ⇒z =C

1

y+C

1

−1⇔ dyC

₁

y+C

₁

−1 =dx (∗)• C

₁

= 0 ⇒ (∗)

cho

^{y =C−x}• C

₁

6= 0 ⇒ (∗)

cho

¹C

₁

ln|C

₁

y+C

₁

−1|=x+C

₂

Ngoai ra

^y=C

la nghi^e.m.

Tom la.i nghi^e.m t^o'ng quat:

^{y=C, y =C−x;}_C¹ln|C

₁

y+C

₁

−1|=x+C

₂

1