1.08k Separable differential equations: dy/dx = f(x)g(y)

9 The variables \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { 3 y } \sin ^ { 2 } 2 x$$ It is given that \(y = 0\) when \(x = 0\).
Solve the differential equation and find the value of \(y\) when \(x = \frac { 1 } { 2 }\).

11 The variables \(y\) and \(\theta\) satisfy the differential equation $$( 1 + y ) ( 1 + \cos 2 \theta ) \frac { d y } { d \theta } = e ^ { 3 y }$$ It is given that \(y = 0\) when \(\theta = \frac { 1 } { 4 } \pi\).
Solve the differential equation and find the exact value of \(\tan \theta\) when \(y = 1\).
If you use the following page to complete the answer to any question, the question number must be clearly shown.

8 The coordinates \(( x , y )\) of a general point of a curve satisfy the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \left( 1 - 2 x ^ { 2 } \right) y$$ for \(x > 0\). It is given that \(y = 1\) when \(x = 1\).
Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).

7 The variables \(x\) and \(t\) satisfy the differential equation $$\mathrm { e } ^ { 3 t } \frac { \mathrm {~d} x } { \mathrm {~d} t } = \cos ^ { 2 } 2 x$$ for \(t \geqslant 0\). It is given that \(x = 0\) when \(t = 0\).

1. Solve the differential equation and obtain an expression for \(x\) in terms of \(t\).
2. State what happens to the value of \(x\) when \(t\) tends to infinity.

7

1. Given that \(y = \ln ( \ln x )\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { x \ln x }$$ The variables \(x\) and \(t\) satisfy the differential equation $$x \ln x + t \frac { \mathrm {~d} x } { \mathrm {~d} t } = 0$$ It is given that \(x = \mathrm { e }\) when \(t = 2\).
2. Solve the differential equation obtaining an expression for \(x\) in terms of \(t\), simplifying your answer.
3. Hence state what happens to the value of \(x\) as \(t\) tends to infinity.

7 The variables \(x\) and \(y\) satisfy the differential equation $$\mathrm { e } ^ { 2 x } \frac { \mathrm {~d} y } { \mathrm {~d} x } = 4 x y ^ { 2 }$$ and it is given that \(y = 1\) when \(x = 0\).
Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).

8 In a certain chemical reaction the amount, \(x\) grams, of a substance is increasing. The differential equation satisfied by \(x\) and \(t\), the time in seconds since the reaction began, is $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k x \mathrm { e } ^ { - 0.1 t }$$ where \(k\) is a positive constant. It is given that \(x = 20\) at the start of the reaction.

1. Solve the differential equation, obtaining a relation between \(x , t\) and \(k\).
2. Given that \(x = 40\) when \(t = 10\), find the value of \(k\) and find the value approached by \(x\) as \(t\) becomes large.

7 The variables \(x\) and \(\theta\) satisfy the differential equation $$x \sin ^ { 2 } \theta \frac { \mathrm {~d} x } { \mathrm {~d} \theta } = \tan ^ { 2 } \theta - 2 \cot \theta$$ for \(0 < \theta < \frac { 1 } { 2 } \pi\) and \(x > 0\). It is given that \(x = 2\) when \(\theta = \frac { 1 } { 4 } \pi\).

1. Show that \(\frac { \mathrm { d } } { \mathrm { d } \theta } \left( \cot ^ { 2 } \theta \right) = - \frac { 2 \cot \theta } { \sin ^ { 2 } \theta }\).
   (You may assume without proof that the derivative of \(\cot \theta\) with respect to \(\theta\) is \(- \operatorname { cosec } ^ { 2 } \theta\).)
2. Solve the differential equation and find the value of \(x\) when \(\theta = \frac { 1 } { 6 } \pi\).

10 A gardener is filling an ornamental pool with water, using a hose that delivers 30 litres of water per minute. Initially the pool is empty. At time \(t\) minutes after filling begins the volume of water in the pool is \(V\) litres. The pool has a small leak and loses water at a rate of \(0.01 V\) litres per minute. The differential equation satisfied by \(V\) and \(t\) is of the form \(\frac { \mathrm { d } V } { \mathrm {~d} t } = a - b V\).

1. Write down the values of the constants \(a\) and \(b\).
2. Solve the differential equation and find the value of \(t\) when \(V = 1000\).
3. Obtain an expression for \(V\) in terms of \(t\) and hence state what happens to \(V\) as \(t\) becomes large.

7 The variables \(x\) and \(\theta\) satisfy the differential equation $$\frac { x } { \tan \theta } \frac { \mathrm {~d} x } { \mathrm {~d} \theta } = x ^ { 2 } + 3$$ It is given that \(x = 1\) when \(\theta = 0\).
Solve the differential equation, obtaining an expression for \(x ^ { 2 }\) in terms of \(\theta\).

11 The variables \(x\) and \(y\) satisfy the differential equation $$x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + y ^ { 2 } + y = 0$$ It is given that \(x = 1\) when \(y = 1\).

1. Solve the differential equation to obtain an expression for \(y\) in terms of \(x\).
2. State what happens to the value of \(y\) when \(x\) tends to infinity. Give your answer in an exact form.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

8 The variables \(x\) and \(y\) satisfy the differential equation $$\mathrm { e } ^ { 4 x } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \cos ^ { 2 } 3 y .$$ It is given that \(y = 0\) when \(x = 2\).
Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).

10 A water tank is in the shape of a cuboid with base area \(40000 \mathrm {~cm} ^ { 2 }\). At time \(t\) minutes the depth of water in the tank is \(h \mathrm {~cm}\). Water is pumped into the tank at a rate of \(50000 \mathrm {~cm} ^ { 3 }\) per minute. Water is leaking out of the tank through a hole in the bottom at a rate of \(600 \mathrm {~cm} ^ { 3 }\) per minute.

1. Show that \(200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 250 - 3 h\).
   
   \includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-17_2723_33_99_22}
2. It is given that when \(t = 0 , h = 50\). Find the time taken for the depth of water in the tank to reach 80 cm . Give your answer correct to 2 significant figures.

6 Use the substitution \(y = v x\) to find the solution of the differential equation $$x \frac { d y } { d x } = y + \sqrt { 9 x ^ { 2 } + y ^ { 2 } }$$ for which \(y = 0\) when \(x = 1\). Give your answer in the form \(\mathrm { y } = \mathrm { f } ( \mathrm { x } )\), where \(\mathrm { f } ( x )\) is a polynomial in \(x\). [10] \includegraphics[max width=\textwidth, alt={}, center]{114ece0d-558d-4c02-8a77-034b3681cff9-10_51_1648_527_246}

6 Use the substitution \(y = v x\) to find the solution of the differential equation $$x \frac { d y } { d x } = y + \sqrt { 9 x ^ { 2 } + y ^ { 2 } }$$ for which \(y = 0\) when \(x = 1\). Give your answer in the form \(\mathrm { y } = \mathrm { f } ( \mathrm { x } )\), where \(\mathrm { f } ( x )\) is a polynomial in \(x\). [10] \includegraphics[max width=\textwidth, alt={}, center]{69c540e1-1dad-45a1-9809-7629d16260e0-10_51_1648_527_246}

2 Use the substitution \(z = x + y\) to find the solution of the differential equation $$\frac { d y } { d x } = \frac { 1 + 3 x + 3 y } { 3 x + 3 y - 1 }$$ for which \(y = 0\) when \(x = 1\). Give your answer in the form \(\operatorname { aln } ( \mathrm { x } + \mathrm { y } ) + \mathrm { b } ( \mathrm { x } - \mathrm { y } ) + \mathrm { c } = 0\), where \(a , b\) and \(c\) are constants to be determined.

6 \includegraphics[max width=\textwidth, alt={}, center]{3e7472a8-df1e-45c4-81fb-e4397bddf5ad-4_182_844_264_653} A particle \(P\) of mass 0.4 kg travels on a horizontal surface along the line \(O A\) in the direction from \(O\) to \(A\). Air resistance of magnitude \(0.1 v \mathrm {~N}\) opposes the motion, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of \(P\) at time \(t \mathrm {~s}\) after it passes through the fixed point \(O\) (see diagram). The speed of \(P\) at \(O\) is \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Assume that the horizontal surface is smooth. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} x } = - \frac { 1 } { 4 }\), where \(x \mathrm {~m}\) is the distance of \(P\) from \(O\) at time \(t \mathrm {~s}\), and hence find the distance from \(O\) at which the speed of \(P\) is zero.
2. Assume instead that the horizontal surface is not smooth and that the coefficient of friction between \(P\) and the surface is \(\frac { 3 } { 40 }\).
   1. Show that \(4 \frac { \mathrm {~d} v } { \mathrm {~d} t } = - ( v + 3 )\).
   2. Hence find the value of \(t\) for which the speed of \(P\) is zero.

4 A particle of mass 0.2 kg moves in a straight line on a smooth horizontal surface. When its displacement from a fixed point on the surface is \(x \mathrm {~m}\), its velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The motion is opposed by a force of magnitude \(\frac { 1 } { 3 v } \mathrm {~N}\).

1. Show that \(3 v ^ { 2 } \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 5\).
2. Find the value of \(v\) when \(x = 7.4\), given that \(v = 4\) when \(x = 0\).

7 A particle \(P\) of mass 0.5 kg moves on a horizontal surface along the straight line \(O A\), in the direction from \(O\) to \(A\). The coefficient of friction between \(P\) and the surface is 0.08 . Air resistance of magnitude \(0.2 v \mathrm {~N}\) opposes the motion, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of \(P\) at time \(t \mathrm {~s}\). The particle passes through \(O\) with speed \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when \(t = 0\).

1. Show that \(2.5 \frac { \mathrm {~d} v } { \mathrm {~d} t } = - ( v + 2 )\) and hence find the value of \(t\) when \(v = 0\).
2. Show that \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 6 \mathrm { e } ^ { - 0.4 t } - 2\), where \(x \mathrm {~m}\) is the displacement of \(P\) from \(O\) at time \(t \mathrm {~s}\), and hence find the distance \(O P\) when \(v = 0\).

3 A particle \(P\) starts from a fixed point \(O\) and moves in a straight line. When the displacement of \(P\) from \(O\) is \(x \mathrm {~m}\), its velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and its acceleration is \(\frac { 1 } { x + 2 } \mathrm {~m} \mathrm {~s} ^ { - 2 }\).

1. Given that \(v = 2\) when \(x = 0\), use integration to show that \(v ^ { 2 } = 2 \ln \left( \frac { 1 } { 2 } x + 1 \right) + 4\).
2. Find the value of \(v\) when the acceleration of \(P\) is \(\frac { 1 } { 4 } \mathrm {~m} \mathrm {~s} ^ { - 2 }\).

7 A particle \(P\) of mass 0.25 kg moves in a straight line on a smooth horizontal surface. \(P\) starts at the point \(O\) with speed \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and moves towards a fixed point \(A\) on the line. At time \(t \mathrm {~s}\) the displacement of \(P\) from \(O\) is \(x \mathrm {~m}\) and the velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A resistive force of magnitude (5-x) N acts on \(P\) in the direction towards \(O\).

1. Form a differential equation in \(v\) and \(x\). By solving this differential equation, show that \(v = 10 - 2 x\).
2. Find \(x\) in terms of \(t\), and hence show that the particle is always less than 5 m from \(O\).

6 A particle \(P\) of mass 0.5 kg moves in a straight line on a smooth horizontal surface. At time \(t \mathrm {~s}\), the displacement of \(P\) from a fixed point on the line is \(x \mathrm {~m}\) and the velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It is given that when \(t = 0 , x = 0\) and \(v = 9\). The motion of \(P\) is opposed by a force of magnitude \(3 \sqrt { } v \mathrm {~N}\).

1. By solving an appropriate differential equation, show that \(v = ( 27 - 9 x ) ^ { \frac { 2 } { 3 } }\).
2. Calculate the value of \(x\) when \(t = 0.5\).

5 A particle \(P\) of mass 0.4 kg moves in a straight line on a horizontal surface and has velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at time \(t \mathrm {~s}\). A horizontal force of magnitude \(k \sqrt { } v \mathrm {~N}\) opposes the motion of \(P\). When \(t = 0 , v = 9\) and when \(t = 2 , v = 4\).

1. Express \(\frac { \mathrm { d } v } { \mathrm {~d} t }\) in terms of \(k\) and \(v\), and hence show that \(v = \frac { 1 } { 4 } ( t - 6 ) ^ { 2 }\).
2. Find the distance travelled by \(P\) in the first 3 seconds of its motion.

6 A cyclist and her bicycle have a total mass of 60 kg . The cyclist rides in a horizontal straight line, and exerts a constant force in the direction of motion of 150 N . The motion is opposed by a resistance of magnitude \(12 v \mathrm {~N}\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the cyclist's speed at time \(t \mathrm {~s}\) after passing through a fixed point \(A\).

1. Show that \(5 \frac { \mathrm {~d} v } { \mathrm {~d} t } = 12.5 - v\).
2. Given that the cyclist passes through \(A\) with speed \(11.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), solve this differential equation to show that \(v = 12.5 - \mathrm { e } ^ { - 0.2 t }\).
3. Express the displacement of the cyclist from \(A\) in terms of \(t\).

4 A small object of mass 0.4 kg is released from rest at a point 8 m above the ground. The object descends vertically and when its downwards displacement from its initial position is \(x \mathrm {~m}\) the object has velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). While the object is moving, a force of magnitude \(0.2 v ^ { 2 } \mathrm {~N}\) opposes the motion.

1. Show that \(v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 10 - 0.5 v ^ { 2 }\).
2. Express \(v\) in terms of \(x\).
3. Find the increase in the value of \(v\) during the final 4 m of the descent of the object.