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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2024

Paper 2

Multiplying
$$
y'+P(x)y=Q(x)y^n
$$
by $y^{-n}$ and setting $z=y^{1-n}$ gives
$$
\frac1{1-n}z'+Pz=Q.
$$
Therefore the <Bernoulli differential equation> becomes
$$
\boxed{z'+(1-n)Pz=(1-n)Q}.
$$

For the stated equation,
$$
\dot x+6t^2x=2te^{-t^3}x^{1/2}.
$$
Set $z=\sqrt x$. Then
$$
\dot z+3t^2z=te^{-t^3}.
$$
The integrating factor is $e^{t^3}$, so
$$
\frac d{dt}(e^{t^3}z)=t.
$$
Hence
$$
e^{t^3}\sqrt{x}=\frac{t^2}{2}+C.
$$
The condition $x(1)=4$ gives $C=2e-\tfrac12$. Consequently
$$
\boxed{x(t)=e^{-2t^3}
\left(\frac{t^2}{2}+2e-\frac12\right)^2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 1A / b / Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 1A / b / Solution

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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 1A / b / Solution

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