# Thread: Family of hyperbolas?

1. ## Family of hyperbolas?

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

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

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

3. I copied the text correctly.

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

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

and of mine

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

I'm not sure how you see the asymptotes $\displaystyle x + y = 0$ and $\displaystyle 2x + y + 1 = 0$ 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 $\displaystyle dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

So

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

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

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

now take the derivative w.r.t y

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

So we get that

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

So we get

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

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

7. @ TheEmptySet - How about his