Show that the substitution \(v = \frac { 1 } { y }\) reduces the differential equation
$$\frac { 2 } { y ^ { 3 } } \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } - \frac { 1 } { y ^ { 2 } } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \frac { 2 } { y ^ { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} x } + \frac { 5 } { y } = 17 + 6 x - 5 x ^ { 2 }$$
to the differential equation
$$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} v } { \mathrm {~d} x } + 5 v = 17 + 6 x - 5 x ^ { 2 }$$
Hence find \(y\) in terms of \(x\), given that when \(x = 0 , y = \frac { 1 } { 2 }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 1\).
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Question 11(e):
Substitution derivative:
Answer Marks
Guidance
\(\frac{dv}{dx} = -\frac{1}{y^2}\frac{dy}{dx}\) B1
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Second derivative:
Answer Marks
Guidance
\(\frac{d^2v}{dx^2} = -\frac{1}{y^2}\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2\frac{d^2y}{dy^2} = -\frac{1}{y^2}\frac{d^2y}{dx^2} + \frac{2}{y^3}\left(\frac{dy}{dx}\right)^2\) M1A1
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Transformed equation:
Answer Marks
Guidance
\(\frac{2}{y^3}\left(\frac{dy}{dx}\right)^2 - \frac{1}{y^2}\frac{d^2y}{dx^2} - \frac{2}{y^2}\frac{dy}{dx} + \frac{5}{y} = 17 + 6x - 5x^2 \sim \frac{d^2v}{dx^2} + 2\frac{dv}{dx} + 5v = 17 + 6x - 5x^2\) A1 (4)
(AG)
Auxiliary equation and CF:
Answer Marks
Guidance
\(m^2 + 2m + 5 = 0 \Rightarrow m = -1 \pm 2i \Rightarrow CF: e^{-x}(A\cos 2x + B\sin 2x)\) M1A1
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Particular Integral:
Answer Marks
Guidance
\(v = px^2 + qx + r \Rightarrow v' = 2px + q \Rightarrow v'' = 2p\) M1
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Equate coefficients: \(5p = -5\), \(4p + 5q = 6\), \(2p + 2q + 5r = 17\) M1
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\(p = -1\), \(q = 2\), \(r = 3\) A1
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General solution:
Answer Marks
Guidance
\(v = e^{-x}(A\cos 2x + B\sin 2x) + 3 + 2x - x^2\) A1
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Applying initial conditions:
Answer Marks
Guidance
When \(x = 0\): \(y = \frac{1}{2} \Rightarrow v = 2\) and \(\frac{dy}{dx} = -1 \Rightarrow \frac{dv}{dx} = 4\) —
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\(2 = A + 3 \Rightarrow A = -1\) B1
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\(v' = e^{-x}(-2A\sin 2x + 2B\cos 2x) - e^{-x}(A\cos 2x + B\sin 2x)\) M1
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\(4 = 2B + 1 + 2 \Rightarrow 2B = 1 \Rightarrow B = \frac{1}{2}\) A1
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Final answer:
Answer Marks
Guidance
\(y = \left[e^{-x}\left(\frac{1}{2}\sin 2x - \cos 2x\right) + 3 + 2x - x^2\right]^{-1}\) A1 (10), Total: 14
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# Question 11(e):
**Substitution derivative:**
$\frac{dv}{dx} = -\frac{1}{y^2}\frac{dy}{dx}$ | B1 | —
**Second derivative:**
$\frac{d^2v}{dx^2} = -\frac{1}{y^2}\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2\frac{d^2y}{dy^2} = -\frac{1}{y^2}\frac{d^2y}{dx^2} + \frac{2}{y^3}\left(\frac{dy}{dx}\right)^2$ | M1A1 | —
**Transformed equation:**
$\frac{2}{y^3}\left(\frac{dy}{dx}\right)^2 - \frac{1}{y^2}\frac{d^2y}{dx^2} - \frac{2}{y^2}\frac{dy}{dx} + \frac{5}{y} = 17 + 6x - 5x^2 \sim \frac{d^2v}{dx^2} + 2\frac{dv}{dx} + 5v = 17 + 6x - 5x^2$ | A1 (4) | (AG)
**Auxiliary equation and CF:**
$m^2 + 2m + 5 = 0 \Rightarrow m = -1 \pm 2i \Rightarrow CF: e^{-x}(A\cos 2x + B\sin 2x)$ | M1A1 | —
**Particular Integral:**
$v = px^2 + qx + r \Rightarrow v' = 2px + q \Rightarrow v'' = 2p$ | M1 | —
Equate coefficients: $5p = -5$, $4p + 5q = 6$, $2p + 2q + 5r = 17$ | M1 | —
$p = -1$, $q = 2$, $r = 3$ | A1 | —
**General solution:**
$v = e^{-x}(A\cos 2x + B\sin 2x) + 3 + 2x - x^2$ | A1 | —
**Applying initial conditions:**
When $x = 0$: $y = \frac{1}{2} \Rightarrow v = 2$ and $\frac{dy}{dx} = -1 \Rightarrow \frac{dv}{dx} = 4$ | — | —
$2 = A + 3 \Rightarrow A = -1$ | B1 | —
$v' = e^{-x}(-2A\sin 2x + 2B\cos 2x) - e^{-x}(A\cos 2x + B\sin 2x)$ | M1 | —
$4 = 2B + 1 + 2 \Rightarrow 2B = 1 \Rightarrow B = \frac{1}{2}$ | A1 | —
**Final answer:**
$y = \left[e^{-x}\left(\frac{1}{2}\sin 2x - \cos 2x\right) + 3 + 2x - x^2\right]^{-1}$ | A1 (10), **Total: 14** | —
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Show that the substitution $v = \frac { 1 } { y }$ reduces the differential equation
$$\frac { 2 } { y ^ { 3 } } \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } - \frac { 1 } { y ^ { 2 } } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \frac { 2 } { y ^ { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} x } + \frac { 5 } { y } = 17 + 6 x - 5 x ^ { 2 }$$
to the differential equation
$$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} v } { \mathrm {~d} x } + 5 v = 17 + 6 x - 5 x ^ { 2 }$$
Hence find $y$ in terms of $x$, given that when $x = 0 , y = \frac { 1 } { 2 }$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = - 1$.
\hfill \mbox{\textit{CAIE FP1 2015 Q11 EITHER}}