1. ## Differential Equations

I'm not sure if I've put this in the correct section, but here goes...
Please may someone help me with this question. I understand how to do (a) but not (b).

The gradient of a curve at the point (x,y) is given by the differential equation
dy/dx = (2-x)/y
a) Find the general solution of dy/dx = (2-x)/y
b) Given that the curve passes through the point (4,2) show that the curve is a circle and find its radius and the coordinates of its centre.

Thank you all
Rachel

2. Originally Posted by racheltllong
I'm not sure if I've put this in the correct section, but here goes...
Please may someone help me with this question. I understand how to do (a) but not (b).

The gradient of a curve at the point (x,y) is given by the differential equation
dy/dx = (2-x)/y
a) Find the general solution of dy/dx = (2-x)/y
b) Given that the curve passes through the point (4,2) show that the curve is a circle and find its radius and the coordinates of its centre.

Thank you all
Rachel
You should have $\frac{y^2}{2} = 2x - \frac{x^2}{2} + C$

$\Rightarrow y^2 = 4x - x^2 + K \Rightarrow (x - 2)^2 + y^2 = A$

where A = K - 4 is just as arbitrary as K and K = 2C is just as arbitrary as C and C is an arbitrary constant.

You can solve for A using the given condition, right?

And you can recognise the equation of a circle. And get the radius and coordinates of centre from that equation, right?

3. Is this really something taught in Pre-Calc ?

4. He probably did not see the calculus' forum, but it's not taught at pre-calc. level.

5. I was about to say, I'm missing out.