# Differential Equations

• Jun 6th 2008, 04:54 AM
racheltllong
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
• Jun 6th 2008, 05:51 AM
mr fantastic
Quote:

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 $\displaystyle \frac{y^2}{2} = 2x - \frac{x^2}{2} + C$

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

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?
• Jun 6th 2008, 06:11 AM
Jonboy
Is this really something taught in Pre-Calc ?
• Jun 6th 2008, 06:14 AM
Krizalid
He probably did not see the calculus' forum, but it's not taught at pre-calc. level.
• Jun 6th 2008, 06:22 AM
Jonboy
I was about to say, I'm missing out.