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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2020

Paper 4

For zero products, put $A=k_1+k'_1+k_2$ and
$$
\lambda_\pm=\frac12\left(A\pm\sqrt{A^2-4k_1k_2}\right).
$$
The initial conditions are $E_0(0)=1$ and $C_0(0)=0$. Eliminating $E_0$ gives
$$
\ddot C_0+A\dot C_0+k_1k_2C_0=0,
\qquad \dot C_0(0)=k_1.
$$
Therefore
$$
C_0(t)=\frac{k_1}{\lambda_+-\lambda_-}
\left(e^{-\lambda_-t}-e^{-\lambda_+t}\right),
$$
and using $E_0=(\dot C_0+(k'_1+k_2)C_0)/k_1$ gives
$$
\boxed{E_0(t)=\frac{(k'_1+k_2-\lambda_-)e^{-\lambda_-t}
-(k'_1+k_2-\lambda_+)e^{-\lambda_+t}}
{\lambda_+-\lambda_-}.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 14B / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 14B / ii / Solution

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Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 14B / ii / Solution

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