# Implicit Form

• Jul 14th 2009, 01:28 PM
stewpot
Implicit Form
Completly stuck and do not have a clue, please help! I need to find in implicit form the general solution of the differential equation:

$\displaystyle \frac{dy}{dx} = 3y^2e^{-2x}\sqrt{8+e^{-2x}}$

This is then followed by finding the corresponding particular solution (in implicit form) that satisfies the initial condition:

y=1/6 when x=0

Many thanks
• Jul 14th 2009, 02:16 PM
skeeter
$\displaystyle \frac{dy}{dx} = 3y^2e^{-2x}\sqrt{8+e^{-2x}}$

$\displaystyle \frac{dy}{y^2} = 3e^{-2x}\sqrt{8+e^{-2x}} \, dx$

$\displaystyle -\frac{1}{y} = -\left(8 + e^{-2x}\right)^{\frac{3}{2}} + C$

can you finish up?
• Jul 15th 2009, 01:25 PM
stewpot

I've come up with this solution to finish it, could you please check it and if it's wrong please show me where!

From where you left me:

8y dy/dx = (e^-2x)^3/2dx

y8=2/x-c

y=4/3 x=0

Many thanks
• Jul 16th 2009, 06:08 AM
skeeter
$\displaystyle \frac{dy}{dx} = 3y^2e^{-2x}\sqrt{8+e^{-2x}}$

$\displaystyle \frac{dy}{y^2} = 3e^{-2x}\sqrt{8+e^{-2x}} \, dx$

$\displaystyle -\frac{1}{y} = -\left(8 + e^{-2x}\right)^{\frac{3}{2}} + C$

$\displaystyle y = \frac{1}{\left(8 + e^{-2x}\right)^{\frac{3}{2}} + C}$

$\displaystyle y(0) = \frac{1}{6}$

$\displaystyle \frac{1}{6} = \frac{1}{27 + C}$

$\displaystyle C = -21$

$\displaystyle y = \frac{1}{\left(8 + e^{-2x}\right)^{\frac{3}{2}} - 21}$