Originally Posted by

**Glitch** **The question:**

Solve the differential equation

$\displaystyle y\sqrt{2x^2 + 3}dy + x\sqrt{4 - y^2}dx = 0$

given that y = 1 when x = 0

**My attempt:**

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

Upon taking the integrals, I got:

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

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.