1. ## Family of hyperbolas?

Show that the following differential equation $(4x + 2y + 1)dx + (3x + 2y + 1)dy = 0$ represents a family of hyperbolas having lines $x + y = 0$ and $2x + y + 1 = 0$ as asymptotes.

EDIT:
As Danny rightly pointed out, the differential equation is actually:
$(4x + 3y + 1)dx + (3x + 2y + 1)dy = 0$

2. Originally Posted by fardeen_gen
Show that the following differential equation $(4x + 2y + 1)dx + (3x + 2y + 1)dy = 0$ represents a family of hyperbolas having lines $x + y = 0$ and $2x + y + 1 = 0$ as asymptotes.
I believe there's a typo here. I think the ODE is $(4x + 3y + 1)dx + (3x + 2y + 1)dy = 0$.

3. I copied the text correctly.

4. I'll let you be the judge. The solution to your problem

$
\frac{2}{\sqrt{7}} \tan^{-1} \frac{5x+4y+2}{\sqrt{7}x} + \ln \left| 8x^2+5(2y+1)x +(2y+1)^2\right| = c
$

and of mine

$
(x+y)(2x+y+1)= c
$
.

I'm not sure how you see the asymptotes $x + y = 0$ and $2x + y + 1 = 0$ coming out in your solution.

5. Can you fill in the details?

6. Originally Posted by fardeen_gen
Can you fill in the details?
The equation is seperable

so $dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

So

$\frac{\partial z}{\partial x}=4x+3y+1$

Take the partial integral w.r.t x and we get

$z(x,y)=2x^2+3xy+x+g(y)$

now take the derivative w.r.t y

$\frac{\partial z}{\partial y}=3x+g'(y)=3x+2y+1$

So we get that

$g'(y)=2y+1 \implies g(y) =y^2+y$

So we get

$2x^2+3xy+x+y^2+y=c \iff 2x^2+xy+x+y^2+2xy+y=c$

$x(2x+y+1)+y(y+2x+1)=c \iff (x+y)(2x+y+1)=c$

7. @ TheEmptySet - How about his