Question

2) (x^2 +y^2) dx + 2xydy =0
Solve the differential equation ​

Answer 1

Step-by-step explanation:

Given, (x

2

+y

2

)dx−2xydy=0

⇒(x

2

+y

2

)dx=2xydy

⇒

dx

dy

=

2xy

x

2

+y

2

.... (i)

Let y=vx

Thus,

dx

dy

=v+x

dx

dv

Thus, v+x

dx

dv

=

2x(vx)

x

2

+(vx)

2

⇒v+x

dx

dv

=

2v

1+v

2

⇒x

dx

dv

=

2v

1+v

2

−v

⇒x

dx

dv

=

2v

1+v

2

−2v

2

⇒x

dx

dv

=

2v

1−v

2

⇒

x

dx

=

1−v

2

2v

dv

⇒

x

dx

−

1−v

2

2v

dv=0 .... (ii)

Integrating both sides, we have

logx+log(1−v

2

)=logC

⇒logx(1−v

2

)=logC

⇒x(1−v

2

)=C

⇒x(1−

x

2

y

2

)=C

⇒x(

x

2

x

2

−y

2

)=C

⇒x

2

−y

2

=Cx

hope it will help you

Answer 2

Answer:

please mark me as branlist