Question

ydx – xdy + log x dx = 0.​

Answer 1

Answer:

ydx=xdy−logxdx

y=x

dx

dy

−logx

⇒

x

logx

=

dx

dy

−1/x

⇒

x

2

logx

=

dx

d

(y/x)

∫

x

2

logx

dx=∫d(y/x)+c

1

∫

x

2

logx

dx=y/x+c

1

let logx=t⇒1/x dx=dt

x=e

t

1/x=e

−t

⇒∫te

−t

dt=−te

−t

−e

−t

=−(

e

t

1+t

)=−

x

(1+logx)

⇒−(

x

1+logx

)=y/x+c

1

⇒1+logx+y+c

1

x=0

put c

1

=−c

⇒1+logx+y=