Implicit Form

**stewpot** · July 14th 2009, 01:28 PM

Completly stuck and do not have a clue, please help! I need to find in implicit form the general solution of the differential equation:

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

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

**skeeter** · July 14th 2009, 02:16 PM

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

$\frac{dy}{y^2} = 3e^{-2x}\sqrt{8+e^{-2x}} \, dx<br />$

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

can you finish up?

**stewpot** · July 15th 2009, 01:25 PM

Hi thanks for your reply, still being emotional about this!

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

**skeeter** · July 16th 2009, 06:08 AM

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

$\frac{dy}{y^2} = 3e^{-2x}\sqrt{8+e^{-2x}} \, dx<br />$

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

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

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

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

$C = -21$

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

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