DH P2 Q3

The height \(x\), of a certain plant, at time \(t\) is proportional to the amount of nutrients it absorbs. The plant loses nutrients at a rate proportional to the height of the plant and absorbs nutrients at a rate inversely proportional to the surface area of the stem of the plant. Taking the cross-section of the stem to be circular with radius \(r=kx\), where \(k\) is a constant, and the surface area of the stem to be \(2\pi rx\),

1. show that \(x\) satisfies the differential equation
   \(\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{p}{{{x}^{2}}}-qx\),
   Where \(p\) and \(q\) are positive constants.[2]
2. For the case \(p=2q\), it is known that \(x=0\) and \(t=0\).
   Solve this differential equation, obtaining an expression for \(x\) in terms of \(t\) and \(q\).[5]
3. Deduce the value of \(x\) as \(t\) becomes very large.[1]