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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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}}