# Differential equation in implicit form

• January 25th 2009, 10:56 AM
bobred
Differential equation in implicit form
I was wondering if someone could give me some idea as where to start with the following question?

Find in implicit form the general solution of the differential equation

$dy/dx\ = -3 y^3 (1 + e^{-3x} )(3x - e^{-3x} + 3)$

Any help would be greatly appreciated

Thanks
• January 25th 2009, 11:29 AM
Jester
Quote:

Originally Posted by bobred
I was wondering if someone could give me some idea as where to start with the following question?

Find in implicit form the general solution of the differential equation

$dy/dx\ = -3 y^3 (1 + e^{-3x} )(3x - e^{-3x} + 3)$

Any help would be greatly appreciated

Thanks

Your equation is separable

$\frac{dy}{y^3} = -3 (1 + e^{-3x} )(3x - e^{-3x} + 3) dx$

expand the RHS and integrate term by term.
• January 25th 2009, 11:34 AM
Opalg
Quote:

Originally Posted by bobred
I was wondering if someone could give me some idea as where to start with the following question?

Find in implicit form the general solution of the differential equation

$dy/dx\ = -3 y^3 (1 + e^{-3x} )(3x - e^{-3x} + 3)$

Divide through by -y^3 and integrate: $-\int\frac{dy}{y^3} = \int 3(1 + e^{-3x} )(3x - e^{-3x} + 3)dx$. The x-integral is easy, because that function is the derivative of $\tfrac12(3x - e^{-3x} + 3)^2$.
• January 25th 2009, 12:51 PM
bobred
Thanks to you both, it's clear now.

James
• January 26th 2009, 05:15 AM
bobred
Hi

I take $-\int\frac{dy}{y^3}$ is the same as $-\int\frac{1}{y^3}dy$ in which case I find the explicit form and solve for y=1/2 and x=0.

But the next part asks for the explicit form which I have had a go at, when I use y=1/2 and x=0 I get a different answer is this correct?

Thanks
• January 26th 2009, 11:01 AM
Opalg
Quote:

Originally Posted by bobred
Hi

I take $-\int\frac{dy}{y^3}$ is the same as $-\int\frac{1}{y^3}dy$ in which case I find the explicit form and solve for y=1/2 and x=0.

But the next part asks for the explicit form which I have had a go at, when I use y=1/2 and x=0 I get a different answer is this correct?

When you do the integration you should get the implicit form of the solution as $y^{-2} = (3x-e^{-3x}+3)^2 +$ const. If the initial condition is that y=1/2 when x=0 then the constant is 0. You can then take the square root of both sides to get the explicit solution as $y = \pm\frac1{3x-e^{-3x}+3}$. Finally, check the initial condition again to see that you need the + sign, not the – sign.
• January 26th 2009, 01:26 PM
bobred
Thanks, I'd been looking at it for ages and going round and round in circles.
• March 15th 2009, 02:48 PM
Confuzed
Quote:

Originally Posted by Opalg
When you do the integration you should get the implicit form of the solution as $y^{-2} = (3x-e^{-3x}+3)^2 +$ const. If the initial condition is that y=1/2 when x=0 then the constant is 0. You can then take the square root of both sides to get the explicit solution as $y = \pm\frac1{3x-e^{-3x}+3}$. Finally, check the initial condition again to see that you need the + sign, not the – sign.

I might be being totally blonde here, but isn't the square root of y^-2 = y^-1?
• March 16th 2009, 07:28 AM
bobred
Reciprocal
Quote:

Originally Posted by Confuzed
I might be being totally blonde here, but isn't the square root of y^-2 = y^-1?

Yes you're right, but then take the reciprocal of both sides.