# Thread: Another differential equation problem

1. ## Another differential equation problem

The question:

Solve the differential equation

$y\sqrt{2x^2 + 3}dy + x\sqrt{4 - y^2}dx = 0$
given that y = 1 when x = 0

My attempt:
$\int {\frac{y dy}{\sqrt{4 - y^2}}} = \int{\frac{x dx}{\sqrt{2x^2 + 3}}$

Upon taking the integrals, I got:

$-\sqrt{4 - y^2}=\frac{\sqrt{2x^2 + 3}}{2} + C$

However solving for y, this gets quite messy, which doesn't seem right considering this is only supposed to be a 2 mark question (practice exam). There's no included answers for me to check. Any assistance would be great.

2. Originally Posted by Glitch
The question:

Solve the differential equation

$y\sqrt{2x^2 + 3}dy + x\sqrt{4 - y^2}dx = 0$
given that y = 1 when x = 0

My attempt:
$\int {\frac{y dy}{\sqrt{4 - y^2}}} = \int{\frac{x dx}{\sqrt{2x^2 + 3}}$ Mr F says: You are missing a negative sign.

Upon taking the integrals, I got:

$-\sqrt{4 - y^2}=\frac{\sqrt{2x^2 + 3}}{2} + C$

However solving for y, this gets quite messy, which doesn't seem right considering this is only supposed to be a 2 mark question (practice exam). There's no included answers for me to check. Any assistance would be great.
Fix the error and use the boundary condition to get C. Now you have a perfectly good implicit solution. Why are you trying to make y the subject?

3. I guess I'm just used to questions expecting a solution for y. I take it that implicit solutions are fine? Thanks.