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IB Mathematics: Applications and Interpretation HL · AHL 9: Differential Equations · AHL 9.17
Differential Equations: exam review Exam review: second paper Answers
Section A: short questions
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- (a) \(\frac{dI}{dt}=kI(1200-I)\)
- (b) \(y=4 x^{2} - \frac{25}{2} + \frac{3}{x}\)
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- (a) \(e^{y}=x^{3}+c\)
- (b) \(y=\ln\left(x^{3}+e^{2}\right)\)
- (c) \(y=2.13\)
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- (a) \(k=\frac{3}{40}=0.0750\)
- (b) shown
- (c) \(9.24\text{ days}\)
- (d) \(46.2\text{ days}\)
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- (a) \(T=25+60e^{-kt}\), \(k=0.101\)
- (b) \(17.7\text{ minutes}\)
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- (a) shown
- (b) \(m=150-120e^{-t/100}\)
- (c) \(87.5\text{ minutes}\)
- (d) \(150\text{ kg}\)
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- (a) \(4\) and \(1\)
- (b) shown
- (c) a curve with a minimum at (0, 0), rising on both sides and staying above \(y=-1\)
- (d) \(y\approx 1.7\) (accept a value within 0.3 of 1.7)
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- (a) \(P=0\) and \(P=80\)
- (b) \(P=80\) is stable
- (c) \(P=40\)
- (d) \(6\)
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- (a) \(\lambda=-1\pm 2 i\ (\approx -1\pm 2.00i)\)
- (b) stable focus; spiral in towards the origin, so \(x\to0\) and \(y\to0\)
- (c) \(\frac{dx}{dt}=2\): clockwise
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- (a) \(\theta(0.2)\approx 0.5216\)
- (b) \(\theta(0.2)\approx 0.5200\)
Section B: long questions
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- (a) \(y(1)\approx 0.297\)
- (b) \(y(1)\approx 0.334\)
- (c) \(y=\frac{1}{4 - \frac{3 x^{2}}{2}}\)
- (d) \(16.5\%\)
- (e) the curve is concave up; use a smaller step length
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- (a) \(x_{n+1}=x_n+h(0.5x_n-0.02x_ny_n),\ y_{n+1}=y_n+h(0.005x_ny_n-0.4y_n)\)
- (b) \(x(1)\approx 105,\ y(1)\approx 35.8\)
- (c) \(x(4)\approx 51.8,\ y(4)\approx 32.5\)
- (d) \(37.8\) (at \(t=2.1\))
- (e) the Euler peaks do grow, but that is Euler error: the exact solution is a repeating cycle with peaks that do not grow, so the statement is not supported; a smaller step length reduces the growth
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- (a) \(A=\begin{pmatrix}4&3\\1&2\end{pmatrix}\)
- (b) \(\lambda=1\) and \(\lambda=5\)
- (c) \(\lambda=1:\ \begin{pmatrix}1\\-1\end{pmatrix}\); \(\lambda=5:\ \begin{pmatrix}3\\1\end{pmatrix}\) (or multiples)
- (d) \(\begin{pmatrix}x\\y\end{pmatrix}=C\begin{pmatrix}1\\-1\end{pmatrix}e^{t}+D\begin{pmatrix}3\\1\end{pmatrix}e^{5 t}\)
- (e) \(x=21 e^{5 t} + e^{t}\); \(y=7 e^{5 t} - e^{t}\)
- (f) unstable node (both eigenvalues positive); see mark scheme for the sketch
- (g) \(y:x=1:3\)
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- (a) \(\frac{dx}{dt}=y,\ \frac{dy}{dt}=- 6 x - 5 y;\ \begin{pmatrix}0&1\\-6&-5\end{pmatrix}\)
- (b) \(x(0.5)\approx 1.12\)
- (c) \(\lambda=-3\text{ or }-2\); stable
- (d) \(x=5 e^{- 2 t} - 2 e^{- 3 t}\)
- (e) \(19.2\%\)
- (f) (i) not stable; (ii) not stable
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- (a) \((3,1)\)
- (b) \(\lambda=-4.732\) or \(-1.268\); stable
- (c) \((3.80,1.10)\ne(4,1)\), so the claim is false
- (d) \(x\to3,\ y\to1\)
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