STEP III 2007 question 8 solution

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STEP III 2007 question 8 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 2007 question 8 solution

(i)

Putting u = x:

$0 + \text{a} (x) + x\text{b} (x) = 0$

Putting u = exp(-x):

$1 - \text{a} (x) + \text{b} (x) = 0$

Adding:

$1 + (x+1)\text{b} (x) = 0 \Rightarrow \text{b} (x) = \frac{-1}{x+1}$

$\therefore \text{a} (x) = \frac{x}{x+1}$

General solution: $u = Ax + Be^{-x} .$

$\displaystyle \frac{\text{d} y}{\text{d} x} + 3y^2 + \frac{x}{1+x} y = \frac{1}{3(1+x)} .$

$\displaystyle y = \frac{1}{3u} \frac{\text{d} u}{\text{d} x} \Rightarrow y^2 = \frac{1}{9u^2} \left(\frac{\text{d} u}{\text{d} x}\right)^2, \quad \frac{\text{d} y}{\text{d} x} = \frac{-1}{3u^2}\frac{\text{d} u}{\text{d} x} + \frac{1}{3u} \frac{\text{d}^2 u}{\text{d} x^2}$

Applying substitution,

$\displaystyle \frac{-1}{3u^2}\left(\frac{\text{d} u}{\text{d} x}\right)^2 + \frac{1}{3u} \frac{\text{d}^2 u}{\text{d} x^2} + \frac{1}{3u^2} \left(\frac{\text{d} u}{\text{d} x}\right)^2 + \frac{x}{1+x} \frac{1}{3u} \frac{\text{d} u}{\text{d} x} = \frac{1}{3(1+x)}$

$\displaystyle \frac{\text{d}^2 u}{\text{d} x^2} + \frac{x}{1+x} \frac{\text{d} u}{\text{d} x} - \frac{1}{1+x} u = 0$

which has general solution $u = Ax + Be^{-x} ,$

i.e.

$\displaystyle \frac{1}{3u} = \frac{1}{3(Ax + Be^{-x})} ,$

$\displaystyle \frac{\text{d} u}{\text{d} x} = A - Be^{-x} ,$

$\displaystyle y = \frac{A - Be^{-x}}{3(Ax + Be^{-x})} .$

y(0) = 0:

$\displaystyle 0 = A - B \Rightarrow A = B$

$\displaystyle \therefore y = \frac{A - Ae^{-x}}{3(Ax + Ae^{-x})} = \frac{1 - e^{-x}}{3(x + e^{-x})}.$

(ii)

$\displaystyle \frac{\text{d} y}{\text{d} x} + y^2 + \frac{x}{1-x} y = \frac{1}{1-x} .$

Applying the substitution $\displaystyle y = \frac{1}{u} \frac{\text{d} u}{\text{d} x}$ gives (detail omitted):

$\displaystyle \frac{\text{d}^2u}{\text{d} x^2} + \frac{x}{1-x} \frac{\text{d} u}{\text{d} x} - \frac{1}{1-x} u = 0$

to which "obvious" solutions are u = x, u = exp x, and so the general solution is u = Ax + Be^x. This gives:

$\displaystyle \frac{1}{u} = \frac{1}{Ax + Be^x} ,\quad \frac{\text{d} u}{\text{d} x} = A + Be^x$

$\displaystyle \therefore y = \frac{A + Be^x}{Ax + Be^x} .$

y(0) = 2:

$\displaystyle 2 = \frac{A + B}{B} \Rightarrow A = B$

$\displaystyle \therefore y = \frac{A + Ae^x}{Ax + Ae^x} = \frac{1 + e^x}{x + e^x} .$

Solution by generalebriety.

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