Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 336 / 2 / ii

An integrating factor explains the second hint and the first-order inner calculation. For

Y^{''} + 2 z Y^{'} = 2 a_{1} z + a_{2} + a_{3} erf z,

e^{z^{2}}

gives

(e^{z^{2}} Y^{'})^{'} = e^{z^{2}} (2 a_{1} z + a_{2} + a_{3} erf z)

. The term

a_{1} z

already produces

2 a_{1} z

, and integrating the remaining terms twice gives

a_{2} E_{0} + a_{3} E_{1}

Y = a_{1} z + a_{2} E_{0} (z) + a_{3} E_{1} (z) + K_{0} + K_{1} erf z .

For the piecewise forcing in the inner problem,

F

replaces

a_{1} z

and provides the continuously matched particular solution for

2 zH (- z)

2 c

and

2 d

then account for the

2 Y_{0}

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