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First order differential equations

Hi,

I am having some problems understanding this topic properly. This is a question which I have attempted and can't see how to get to the answer given by the book (dydx=xy \frac{dy}{dx} = \frac{-x}{y}).

By eliminating the arbitrary constant A, find a first order differential equation that is equivalent to x2+y2=A x^2 + y^2 = A

First, I differentiated all terms with respect to x:

2x+2ydydx=Adydx 2x + 2y\frac{dy}{dx} = A\frac{dy}{dx}

Given that Adydx=(x2+y2)dydxA\frac{dy}{dx} = (x^2 + y^2)\frac{dy}{dx},

2x+2ydydx=x2dydx+y2dydx 2x + 2y\frac{dy}{dx}= x^2\frac{dy}{dx} + y^2\frac{dy}{dx}

2x=x2dydx+y2dydx2ydydx2x = x^2\frac{dy}{dx} + y^2\frac{dy}{dx} -2y\frac{dy}{dx}

2x=dydx(x2+y22y)2x = \frac{dy}{dx}(x^2+y^2-2y)

dydx=2xx2+y22y \frac{dy}{dx} = \frac{2x}{x^2+y^2-2y}

Any help on this would be appreciated because I am at a loss at the moment..!
Reply 1
Original post by jonnburton
Hi,

I am having some problems understanding this topic properly. This is a question which I have attempted and can't see how to get to the answer given by the book (dydx=xy \frac{dy}{dx} = \frac{-x}{y}).

By eliminating the arbitrary constant A, find a first order differential equation that is equivalent to x2+y2=A x^2 + y^2 = A

First, I differentiated all terms with respect to x:

2x+2ydydx=Adydx 2x + 2y\frac{dy}{dx} = A\frac{dy}{dx}

Given that Adydx=(x2+y2)dydxA\frac{dy}{dx} = (x^2 + y^2)\frac{dy}{dx},

2x+2ydydx=x2dydx+y2dydx 2x + 2y\frac{dy}{dx}= x^2\frac{dy}{dx} + y^2\frac{dy}{dx}

2x=x2dydx+y2dydx2ydydx2x = x^2\frac{dy}{dx} + y^2\frac{dy}{dx} -2y\frac{dy}{dx}

2x=dydx(x2+y22y)2x = \frac{dy}{dx}(x^2+y^2-2y)

dydx=2xx2+y22y \frac{dy}{dx} = \frac{2x}{x^2+y^2-2y}

Any help on this would be appreciated because I am at a loss at the moment..!


Isn't A just a constant? So when you differentiate in your first step, it just disappears, and rearranging the resulting line gives the answer.

I think the question is a bit misleading, though, talking about "eliminating A", as that usually means getting rid of it using two equations, as you did.
Reply 2
Original post by MAD Phil
Isn't A just a constant? So when you differentiate in your first step, it just disappears, and rearranging the resulting line gives the answer.

I think the question is a bit misleading, though, talking about "eliminating A", as that usually means getting rid of it using two equations, as you did.


What you say makes sense to me but in the worked example from the book they didn't do it like this: they kept the 'A'...

I tried letting A disappear but still came to the same solution:

x2+y2=Ax^2 +y^2 = A

2x+2ydydx=0 2x + 2y \frac{dy}{dx} = 0

2x+2ydydx=(x2+y2)dydx 2x + 2y\frac{dy}{dx} = (x^2 + y^2) \frac{dy}{dx}

2x=x2dydx+y2dydx2ydydx 2x = x^2\frac{dy}{dx} + y^2\frac{dy}{dx} -2y\frac{dy}{dx}

dydx=2xx2+y22y\frac{dy}{dx} = \frac{2x}{x^2 + y^2 -2y}
Reply 3
Original post by jonnburton
What you say makes sense to me but in the worked example from the book they didn't do it like this: they kept the 'A'...

I tried letting A disappear but still came to the same solution:

x2+y2=Ax^2 +y^2 = A

2x+2ydydx=0 2x + 2y \frac{dy}{dx} = 0

2x+2ydydx=(x2+y2)dydx 2x + 2y\frac{dy}{dx} = (x^2 + y^2) \frac{dy}{dx} <-----?

2x=x2dydx+y2dydx2ydydx 2x = x^2\frac{dy}{dx} + y^2\frac{dy}{dx} -2y\frac{dy}{dx}

dydx=2xx2+y22y\frac{dy}{dx} = \frac{2x}{x^2 + y^2 -2y}


How did you get that?

Instead you should have 2ydydx=2x2y\frac{dy}{dx}=-2x.
Reply 4
Original post by BabyMaths
How did you get that?

Instead you should have 2ydydx=2x2y\frac{dy}{dx}=-2x.


I thought I was supposed to substitute the A value back into the given equation.

i.e. I thought that since A=(x2+y2) A = (x^2 + y^2), I should substitute the differential of A, (x2+y2)dydx(x^2 + y^2)\frac{dy}{dx}



I was under the impression this was what they did in the worked example in my book, when finding a differential equation for x2+y2=Ay x^2 + y^2 = Ay

First they differentiate with respect to x:

2x+2ydydx=Adydx 2x + 2y\frac{dy}{dx} = A\frac{dy}{dx}

Then multiply by y:
2xy+2y2dydx=Aydydx 2xy + 2y^2\frac{dy}{dx} = Ay\frac{dy}{dx}

And then using the given equation:
2xy+2y2dydx=(x2+y2)dydx 2xy + 2y^2\frac{dy}{dx} = (x^2 + y^2)\frac{dy}{dx}
Reply 5
Original post by jonnburton
I thought I was supposed to substitute the A value back into the given equation.

i.e. I thought that since A=(x2+y2) A = (x^2 + y^2), I should substitute the differential of A, (x2+y2)dydx(x^2 + y^2)\frac{dy}{dx}



I was under the impression this was what they did in the worked example in my book, when finding a differential equation for x2+y2=Ay x^2 + y^2 = Ay

First they differentiate with respect to x:

2x+2ydydx=Adydx 2x + 2y\frac{dy}{dx} = A\frac{dy}{dx}

Then multiply by y:
2xy+2y2dydx=Aydydx 2xy + 2y^2\frac{dy}{dx} = Ay\frac{dy}{dx}

And then using the given equation:
2xy+2y2dydx=(x2+y2)dydx 2xy + 2y^2\frac{dy}{dx} = (x^2 + y^2)\frac{dy}{dx}


Yes. But differentiating Ay with respect to x gives A dy/dx, while differentiating A gives 0 (as BabyMaths says).
(edited 10 years ago)
Reply 6
OK thanks Mad Phil, I think I'm gradually getting to grips with this now...

Original post by MAD Phil
Yes. But differentiating Ay with respect to x gives A dy/dx, while differentiating A gives 0 (as BabyMaths says).
Reply 7
Original post by jonnburton
OK thanks Mad Phil, I think I'm gradually getting to grips with this now...


That's what TSR is for! Soon you'll be helping other people with problems like this!
Reply 8
Original post by MAD Phil
That's what TSR is for! Soon you'll be helping other people with problems like this!


Yeah, hopefully!

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