Differential Equations

To solve the given differential equation, we start by analyzing the equation:
(x−y−1)dy=(x+y+1)‌dx
This can be rewritten in differential form as:
‌

dy

dx

=‌

x+y+1

x−y−1

This expression can also be represented as:
‌

dy

dx

=‌

x+(y+1)

x−(y+1)

Introduce a substitution where y+1=z. Thus, we have:
‌

dy

dx

=‌

dz

dx

After substitution, the equation becomes:
‌

dz

dx

=‌

x+z

x−z

Next, we use another substitution z=vx. Therefore, we differentiate z with respect to x :
‌

dz

dx

=v+x‌

dv

dx

Substituting this back into our equation results in:
v+x‌

dv

dx

=‌

1+v

1−v

Reorganize terms and separate variables:
x‌

dv

dx

=‌

1+v−v+v2

1−v

This simplifies to:
(‌

1−v

1+v2

)dv=‌

dx

x

Integrating both sides, we have:
∫(‌

1

1+v2

−‌

v

1+v2

)dv=∫‌

dx

x

The integrals yield:

tan−1v−‌

1

2

‌log‌|1+v2|=log‌x+C
Substitute back v=‌

z

x

:
tan−1‌

z

x

−‌

1

2

‌log‌|‌

z2

x2

|=log‌x+C
Simplifying using z=y+1 :
tan−1(‌

y+1

x

)−‌

1

2

‌log‌|‌

x2+(y+1)2

x2

|−log‌x=C
Finally, this simplifies to:
tan−1(‌

y+1

x

)−‌

1

2

‌log‌|x2+y2+2y+1|=C