Question

answered my question​

Answer 1

Step-by-step explanation:

y−x

dx

dy

=x+y

dx

dy

Substitute

y=vx⇒

dx

dy

=v+x

dx

dv

∴−x(x

dx

dv

+v)+xv=x+x(x

dx

dv

+v)v

⇒

dx

dv

=

x(v+1)

−v

2

−1

⇒

−v

2

−1

v+1

dx

dv

=

x

1

Integrating both sides w.r.t x we get

∫

−v

2

−1

v+1

dx

dv

dx=∫

x

1

dx

⇒tan

−1

v−

2

1

log(v

2

+1)=logx+c⇒kx=e

−

x

y