Question

dy/dx 3x^2+xy-y^2+4x-2y+1=0​

Answer 1

Answer:

dy/dx 3x^2+xy-y^2+4x-2y+1=0

Step-by-step explanation:

2x+y+1)dx+(4x+2y−1dy)=0

Solution:

dx

dy

=−

4x+2y−1

2x+y+1

Let, v=2x+y

dx

dv

=2+

dx

dy

dx

dv

=2−

2v−1

v+1

=

2v−1

4v−2−v−1

dx

dv

=

2v−1

3v−3

3v−3

2v−1

dv=dx −−−−−(i)

3v−3

2v−1

=

3

2

+

3v−3

1

Integrating both sides,

or,

3

2

∫dv+∫

3v−3

1

=∫dx

or,

3

2

v+

3

1

log(3v−3)=x+c

or, 2v+log(3v−3)=3x+3c

or, 2(2x+y)+log(3(2x+y)−3)=3x+c

′

or, 4x+2y+log(6x+3y−3)=3x+c

′

or, x+2y+log(6x+3y−3)=k

Hence, the solution is x+2y+log(6x+3y−3)=k.

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