Solve via substitution then back-substitute

7

1. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 10 y = \mathrm { e } ^ { 2 t }$$ giving your answer in the form \(y = \mathrm { f } ( t )\).
2. Given that \(x = t ^ { \frac { 1 } { 2 } } , x > 0 , t > 0\) and \(y\) is a function of \(x\), show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t }$$ (5 marks)
3. Hence show that the substitution \(x = t ^ { \frac { 1 } { 2 } }\) transforms the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 12 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 40 x ^ { 3 } y = 4 x ^ { 3 } \mathrm { e } ^ { 2 x ^ { 2 } }$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 10 y = \mathrm { e } ^ { 2 t }$$ (2 marks)
4. Hence write down the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 12 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 40 x ^ { 3 } y = 4 x ^ { 3 } \mathrm { e } ^ { 2 x ^ { 2 } }$$ (l mark)

6

1. Show that the substitution $$u = \frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y$$ transforms the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = \mathrm { e } ^ { - 2 x }$$ into $$\frac { \mathrm { d } u } { \mathrm {~d} x } + 2 u = \mathrm { e } ^ { - 2 x }$$ (4 marks)
2. By using an integrating factor, or otherwise, find the general solution of $$\frac { \mathrm { d } u } { \mathrm {~d} x } + 2 u = \mathrm { e } ^ { - 2 x }$$ giving your answer in the form \(u = \mathrm { f } ( x )\).
3. Hence find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = \mathrm { e } ^ { - 2 x }$$ giving your answer in the form \(y = \mathrm { g } ( x )\).

7

1. Given that \(x = t ^ { \frac { 1 } { 2 } } , x > 0 , t > 0\) and \(y\) is a function of \(x\), show that:
   1. \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 t ^ { \frac { 1 } { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} t }\);
      (2 marks)
   2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t }\).
      (3 marks)
2. Hence show that the substitution \(x = t ^ { \frac { 1 } { 2 } }\) transforms the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 3 } y = 12 x ^ { 5 }$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = 3 t$$ (2 marks)
3. Hence find the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 3 } y = 12 x ^ { 5 }$$ giving your answer in the form \(y = \mathrm { f } ( x )\).

6 A differential equation is given by $$\left( x ^ { 3 } + 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - 3 x ^ { 2 } \frac { d y } { d x } = 2 - 4 x ^ { 3 }$$

1. Show that the substitution $$u = \frac { \mathrm { d } y } { \mathrm {~d} x } - 2 x$$ transforms this differential equation into $$\left( x ^ { 3 } + 1 \right) \frac { \mathrm { d } u } { \mathrm {~d} x } = 3 x ^ { 2 } u$$ (4 marks)
2. Hence find the general solution of the differential equation $$\left( x ^ { 3 } + 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = 2 - 4 x ^ { 3 }$$ giving your answer in the form \(y = \mathrm { f } ( x )\). \(7 \quad\) The curve \(C _ { 1 }\) is defined by \(r = 2 \sin \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }\). The curve \(C _ { 2 }\) is defined by \(r = \tan \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }\).
3. Find a cartesian equation of \(C _ { 1 }\).
4. 1. Prove that the curves \(C _ { 1 }\) and \(C _ { 2 }\) meet at the pole \(O\) and at one other point, \(P\), in the given domain. State the polar coordinates of \(P\).
   2. The point \(A\) is the point on the curve \(C _ { 1 }\) at which \(\theta = \frac { \pi } { 4 }\). The point \(B\) is the point on the curve \(C _ { 2 }\) at which \(\theta = \frac { \pi } { 4 }\). Determine which of the points \(A\) or \(B\) is further away from the pole \(O\), justifying your answer.
   3. Show that the area of the region bounded by the arc \(O P\) of \(C _ { 1 }\) and the arc \(O P\) of \(C _ { 2 }\) is \(a \pi + b \sqrt { 3 }\), where \(a\) and \(b\) are rational numbers.

7 A differential equation is given by $$\sin ^ { 2 } x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \sin x \cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 \sin ^ { 4 } x \cos x , \quad 0 < x < \pi$$

1. Show that the substitution $$y = u \sin x$$ where \(u\) is a function of \(x\), transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } u } { \mathrm {~d} x ^ { 2 } } + u = \sin 2 x$$
2. Hence find the general solution of the differential equation $$\sin ^ { 2 } x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \sin x \cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 \sin ^ { 4 } x \cos x$$ giving your answer in the form \(y = \mathrm { f } ( x )\).
   (6 marks)

6 A differential equation is given by $$4 \sqrt { x ^ { 5 } } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 2 \sqrt { x } ) y = \sqrt { x } ( \ln x ) ^ { 2 } + 5 , \quad x > 0$$

1. Show that the substitution \(x = \mathrm { e } ^ { 2 t }\) transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 2 y = 4 t ^ { 2 } + 5 \mathrm { e } ^ { - t }$$
2. Hence find the general solution of the differential equation $$4 \sqrt { x ^ { 5 } } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 2 \sqrt { x } ) y = \sqrt { x } ( \ln x ) ^ { 2 } + 5 , \quad x > 0$$
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5

1. Express \(\frac { 1 } { ( 1 + x ) ( 2 + x ) }\) in the form \(\frac { A } { 1 + x } + \frac { B } { 2 + x }\), where \(A\) and \(B\) are integers.
2. Use the substitution \(u = \frac { \mathrm { d } y } { \mathrm {~d} x }\) to solve the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \frac { 1 } { ( 1 + x ) ( 2 + x ) } \frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 + x } { 1 + x }$$ given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 4\) when \(x = 0\). Give your answer in the form \(y = \mathrm { f } ( x )\).
   [0pt] [11 marks]

1. The concentration of a drug in the bloodstream of a patient, \(t\) hours after the drug has been administered, where \(t \leqslant 6\), is modelled by the differential equation

$$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } C } { \mathrm {~d} t ^ { 2 } } - 5 t \frac { \mathrm {~d} C } { \mathrm {~d} t } + 8 C = t ^ { 3 }$$ where \(C\) is measured in micrograms per litre.

1. Show that the transformation \(t = \mathrm { e } ^ { x }\) transforms equation (I) into the equation $$\frac { \mathrm { d } ^ { 2 } C } { \mathrm {~d} x ^ { 2 } } - 6 \frac { \mathrm {~d} C } { \mathrm {~d} x } + 8 C = \mathrm { e } ^ { 3 x }$$
2. Hence find the general solution for the concentration \(C\) at time \(t\) hours. Given that when \(t = 6 , C = 0\) and \(\frac { \mathrm { d } C } { \mathrm {~d} t } = - 36\)
3. find the maximum concentration of the drug in the bloodstream of the patient.

1. A particle \(P\) moves along a straight line.

At time \(t\) minutes, the displacement, \(x\) metres, of \(P\) from a fixed point \(O\) on the line is modelled by the differential equation $$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 2 t \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 x + 16 t ^ { 2 } x = 4 t ^ { 3 } \sin 2 t$$

1. Show that the transformation \(x =\) ty transforms equation (I) into the equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 16 y = 4 \sin 2 t$$
2. Hence find a general solution for the displacement of \(P\) from \(O\) at time \(t\) minutes.

1. The motion of a particle \(P\) along the \(x\)-axis is modelled by the differential equation

$$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 2 t ( t + 1 ) \frac { \mathrm { d } x } { \mathrm {~d} t } + 2 ( t + 1 ) x = 8 t ^ { 3 } \mathrm { e } ^ { t }$$ where \(P\) has displacement \(x\) metres from the origin \(O\) at time \(t\) minutes, \(t > 0\)

1. Show that the transformation \(x = t u\) transforms the differential equation (I) into the differential equation $$\frac { \mathrm { d } ^ { 2 } u } { \mathrm {~d} t ^ { 2 } } - 2 \frac { \mathrm {~d} u } { \mathrm {~d} t } = 8 \mathrm { e } ^ { t }$$ Given that \(P\) is at \(O\) when \(t = \ln 3\) and when \(t = \ln 5\)
2. determine the particular solution of the differential equation (I)

1. A vibrating spring, fixed at one end, has an external force acting on it such that the centre of the spring moves in a straight line. At time \(t\) seconds, \(t \geqslant 0\), the displacement of the centre \(C\) of the spring from a fixed point \(O\) is \(x\) micrometres.

The displacement of \(C\) from \(O\) is modelled by the differential equation $$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 2 t \frac { \mathrm {~d} x } { \mathrm {~d} t } + \left( 2 + t ^ { 2 } \right) x = t ^ { 4 }$$

1. Show that the transformation \(x = t v\) transforms equation (I) into the equation $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} t ^ { 2 } } + v = t$$
2. Hence find the general equation for the displacement of \(C\) from \(O\) at time \(t\) seconds.
3. 1. State what happens to the displacement of \(C\) from \(O\) as \(t\) becomes large.
   2. Comment on the model with reference to this long term behaviour.

5

1. A differential equation is given by $$\left( x ^ { 2 } - 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = x ^ { 2 } + 1$$ Show that the substitution $$u = \frac { \mathrm { d } y } { \mathrm {~d} x } + x$$ transforms this differential equation into $$\frac { \mathrm { d } u } { \mathrm {~d} x } = \frac { 2 x u } { x ^ { 2 } - 1 }$$ (4 marks)
2. Find the general solution of $$\frac { \mathrm { d } u } { \mathrm {~d} x } = \frac { 2 x u } { x ^ { 2 } - 1 }$$ giving your answer in the form \(u = \mathrm { f } ( x )\).
3. Hence find the general solution of the differential equation $$\left( x ^ { 2 } - 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = x ^ { 2 } + 1$$ giving your answer in the form \(y = \mathrm { g } ( x )\).