FP2 June 2010 Q7
7.
(a) Show that the transformation \(z = y^{\frac{1}{2}}\) transforms the differential equation \[\frac{\mathrm{d}y}{\mathrm{d}x} - 4y\tan x = 2y^{\frac{1}{2}} \qquad \text{(I)}\] into the differential equation \[\frac{\mathrm{d}z}{\mathrm{d}x} - 2z\tan x = 1 \qquad \text{(II)}\] (5)
(b) Solve the differential equation (II) to find \(z\) as a function of \(x\). (6)
(c) Hence obtain the general solution of the differential equation (I). (1)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}z} \cdot \dfrac{\mathrm{d}z}{\mathrm{d}x}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}z} = 2z\) so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2z \cdot \dfrac{\mathrm{d}z}{\mathrm{d}x}\) | M1 M1 A1 |
| Substituting to get \(2z \cdot \dfrac{\mathrm{d}z}{\mathrm{d}x} - 4z^2\tan x = 2z\) and thus \(\dfrac{\mathrm{d}z}{\mathrm{d}x} - 2z\tan x = 1\) * | M1 A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\text{I.F.} = \mathrm{e}^{\int -2\tan x\,\mathrm{d}x} = \mathrm{e}^{2\ln\cos x} = \cos^2 x\) | M1 A1 |
| \(\therefore \dfrac{\mathrm{d}}{\mathrm{d}x}(z\cos^2 x) = \cos^2 x \quad \therefore z\cos^2 x = \displaystyle\int \cos^2 x\,\mathrm{d}x\) | M1 |
| \(\therefore z\cos^2 x = \displaystyle\int \tfrac{1}{2}(\cos 2x + 1)\,\mathrm{d}x = \tfrac{1}{4}\sin 2x + \tfrac{1}{2}x + c\) | M1 A1 |
| \(\therefore z = \tfrac{1}{2}\tan x + \tfrac{1}{2}x\sec^2 x + c\sec^2 x\) | A1 |
| (6) |
| Scheme | Marks |
|---|---|
| \(\therefore y = \left(\tfrac{1}{2}\tan x + \tfrac{1}{2}x\sec^2 x + c\sec^2 x\right)^2\) | B1ft |
| (1) | |
| (12 marks) |