Thread: 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



Any help would be greatly appreciated

Thanks

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



Any help would be greatly appreciated

Thanks

Your equation is separable



expand the RHS and integrate term by term.

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

Divide through by -y^3 and integrate: . The x-integral is easy, because that function is the derivative of .

Thanks to you both, it's clear now.

James

Hi

I take is the same as 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

Originally Posted by bobred

Hi

I take is the same as 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 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 . Finally, check the initial condition again to see that you need the + sign, not the – sign.

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

Originally Posted by Opalg

When you do the integration you should get the implicit form of the solution as 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 . 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?

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.