# Math Help - Differential equation in implicit form

1. ## 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

2. 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

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

expand the RHS and integrate term by term.

3. 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 $\tfrac12(3x - e^{-3x} + 3)^2$.

4. Thanks to you both, it's clear now.

James

5. 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

6. 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.

7. Thanks, I'd been looking at it for ages and going round and round in circles.

8. 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?

9. ## Reciprocal

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.