DHS 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
   
   \(\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{p}{x^{2}}-qx\),
   
   where \(p\) and \(q\) are positive constants.[2]
   For the case \({p = 2q}\), it is known that \({x = 0}\) and \({t = 0}\).
2. 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]