Separable variables - standard (requires integration by parts)

7 The variables \(x\) and \(y\) are related by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 x \mathrm { e } ^ { 3 x } } { y ^ { 2 } } .$$ It is given that \(y = 2\) when \(x = 0\). Solve the differential equation and hence find the value of \(y\) when \(x = 0.5\), giving your answer correct to 2 decimal places.

1. (a) Find \(\int x ^ { 2 } \cos 2 x d x\) (b) Hence solve the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( \frac { t \cos t } { y } \right) ^ { 2 }$$ giving your answer in the form \(y ^ { n } = \mathrm { f } ( t )\) where \(n\) is an integer.

8. (a) Given that \(y = 1\) at \(x = 0\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 x y ^ { \frac { 1 } { 3 } } } { \mathrm { e } ^ { 2 x } } \quad y \geqslant 0$$ giving your answer in the form \(y ^ { 2 } = \mathrm { g } ( x )\).
(b) Hence find the equation of the horizontal asymptote to the curve with equation \(y ^ { 2 } = \mathrm { g } ( x )\). \includegraphics[max width=\textwidth, alt={}, center]{960fe82f-c180-422c-b409-a5cdc5fae924-27_2644_1840_118_111} \includegraphics[max width=\textwidth, alt={}, center]{960fe82f-c180-422c-b409-a5cdc5fae924-29_2646_1838_121_116}

11 The gradient function of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 x ^ { 2 } \ln x } { \mathrm { e } ^ { 3 y } }\).
The curve passes through the point (e, 1).

1. Find the equation of this curve, giving your answer in the form \(\mathrm { e } ^ { 3 y } = \mathrm { f } ( x )\).
2. Show that, when \(x = \mathrm { e } ^ { 2 }\), the \(y\)-coordinate of this curve can be written as \(y = a + \frac { 1 } { 3 } \ln \left( b \mathrm { e } ^ { 3 } + c \right)\), where \(a , b\) and \(c\) are constants to be determined.

7 Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ^ { 2 } x \sin 3 x$$ given that \(y = 1\) when \(x = \frac { \pi } { 6 }\). Give your answer in the form \(y = \frac { 9 } { \mathrm { f } ( x ) }\).

12 The gradient function of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ^ { 2 } \sin 2 x } { 2 \cos ^ { 2 } 4 y - 1 }\).

1. Find an equation for the curve in the form \(\mathrm { f } ( y ) = g ( x )\). The curve passes through the point \(\left( \frac { 1 } { 4 } \pi , \frac { 1 } { 12 } \pi \right)\).
2. Find the smallest positive value of \(y\) for which \(x = 0\). \section*{END OF QUESTION PAPER}