Question

solve I.V.P (3xy + y²) dx + (x² + xy) dy = 0 , y(1) = 1​

Answer 1

Step-by-step explanation:

x

2

+xy)

dx

dy

+y

2

+3xy=0

Substituting y=vx⇒

dx

dy

=v+x

dx

dv

(x

2

+x

2

v)(x

dx

dv

+v)+x

2

v

2

+3x

2

v=0

⇒x

2

(x

dx

dv

+2v

2

+(x

dx

dv

+4)v)=0

⇒

dx

dv

=−

x(v+1)

2v

2

+2v

=

x(v+1)

2(v+2)v

⇒

v(v+2)

dx

dv

(v+1)

=−

x

2

Integrating both sides w.r.t x, we get

∫

v(v+2)

dx

dv

(v+1)

dx=∫−

x

2

dx

⇒

2

1

log(v+2)+

2

1

logv=−2logx+c

⇒

2

1

log(

x

y

+2)+

2

1

log

x

y

=−2logx+c

Now y(1)=1⇒x

2

y(2x+y)=3