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

7 Solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \sec y\) given that \(y = \frac { \pi } { 6 }\) when \(x = 4\) giving your answer in the form \(y = \mathrm { f } ( x )\).

10 A cyclist and her bicycle have a combined mass of 90 kg and she is riding along a straight horizontal road. She is working at a constant power of 75 W . At time \(t\) seconds her speed is \(v \mathrm {~ms} ^ { - 1 }\) and the resistance to motion is \(k v \mathrm {~N}\), where \(k\) is a constant.

1. Given that the steady speed at which the cyclist can move is \(10 \mathrm {~ms} ^ { - 1 }\), show that \(k = \frac { 3 } { 4 }\).
2. Show that $$\frac { 25 } { v } - \frac { v } { 4 } = 30 \frac { \mathrm {~d} v } { \mathrm {~d} t } .$$
3. Find the time taken for the cyclist to accelerate from a speed of \(3 \mathrm {~ms} ^ { - 1 }\) to a speed of \(7 \mathrm {~ms} ^ { - 1 }\).

12 A particle of mass 0.6 kg is projected vertically upwards from horizontal ground, with initial speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The upwards velocity at any instant is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and the displacement is \(x \mathrm {~m}\). Air resistance is modelled by a force \(0.024 v ^ { 2 } \mathrm {~N}\) acting downwards.

1. Show that \(v\) and \(x\) satisfy the differential equation $$v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 10 - 0.04 v ^ { 2 } .$$
2. Find the value of \(u\) if the maximum height reached is 50 m .

6 Find the solution of the differential equation $$x y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = x + 1$$ given that \(y = 3\) when \(x = 1\). Give your answer in the form \(y = \mathrm { f } ( x )\).

10

1. Using partial fractions, find the general solution of the differential equation $$2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = y - y ^ { 3 } \text { for } 0 < y < 1$$ giving your solution in the form \(y = \mathrm { f } ( x )\).
2. Determine \(\lim _ { x \rightarrow - \infty } \mathrm { f } ( x )\) and \(\lim _ { x \rightarrow + \infty } \mathrm { f } ( x )\).

10 A cyclist and her bicycle have a combined mass of 90 kg and she is riding along a straight horizontal road. She is working at a constant power of 75 W . At time \(t\) seconds her speed is \(v \mathrm {~ms} ^ { - 1 }\) and the resistance to motion is \(k v \mathrm {~N}\), where \(k\) is a constant.

1. Given that the steady speed at which the cyclist can move is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), show that \(k = \frac { 3 } { 4 }\).
2. Show that $$\frac { 25 } { v } - \frac { v } { 4 } = 30 \frac { \mathrm {~d} v } { \mathrm {~d} t } .$$
3. Find the time taken for the cyclist to accelerate from a speed of \(3 \mathrm {~ms} ^ { - 1 }\) to a speed of \(7 \mathrm {~ms} ^ { - 1 }\).

7 Solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \sec y\) given that \(y = \frac { \neq } { 6 }\) when \(x = 4\) giving your answer in the form \(y = \mathrm { f } ( x )\).

10

1. 1. By writing \(\sec x = \frac { 1 } { \cos x }\), prove that $$\frac { \mathrm { d } } { \mathrm {~d} x } ( \sec x ) = \sec x \tan x .$$
   2. Deduce that \(y = \sec x\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y \sqrt { y ^ { 2 } - 1 } , \quad 0 \leq x < \frac { 1 } { 2 } \pi .$$
2. A curve lies in the first quadrant of the cartesian plane with origin \(O\) as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{85043199-527d-4105-aa0b-c913dec0e35b-4_707_698_845_685} The normal to the curve at the point \(P ( x , y )\) meets the \(x\)-axis at the point \(Q\). The angle between \(O P\) and the \(x\)-axis is \(u\), and the angle between \(Q P\) and the \(x\)-axis is \(v\).
   1. If $$\tan v = \tan ^ { 2 } u$$ obtain a differential equation satisfied by the curve.
   2. The curve passes through the point \(( 2,1 )\). By solving the differential equation, find an equation for the curve in the implicit form $$\mathrm { F } ( x , y ) = C ,$$ where \(C\) is a constant that should be determined.

\includegraphics{figure_9} A container in the shape of a cuboid has a square base of side \(x\) and a height of \((10 - x)\). It is given that \(x\) varies with time, \(t\), where \(t > 0\). The container decreases in volume at a rate which is inversely proportional to \(t\). When \(t = \frac{1}{10}\), \(x = \frac{1}{2}\) and the rate of decrease of \(x\) is \(\frac{20}{37}\).

1. Show that \(x\) and \(t\) satisfy the differential equation $$\frac{dx}{dt} = \frac{-1}{2t(20x - 3x^2)}$$ [5]
2. Solve the differential equation, obtaining an expression for \(t\) in terms of \(x\). [6]

The variables \(x\) and \(y\) satisfy the differential equation $$(1 - \cos x)\frac{dy}{dx} = y \sin x.$$ It is given that \(y = 4\) when \(x = \pi\).

1. Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\). [6]
2. Sketch the graph of \(y\) against \(x\) for \(0 < x < 2\pi\). [1]

Given that \(x = 1\) when \(t = 0\), solve the differential equation $$\frac{dx}{dt} = \frac{1}{x} - \frac{x}{4},$$ obtaining an expression for \(x^2\) in terms of \(t\). [7]

The population of a country at time \(t\) years is \(N\) millions. At any time, \(N\) is assumed to increase at a rate proportional to the product of \(N\) and \((1 - 0.01N)\). When \(t = 0\), \(N = 20\) and \(\frac{dN}{dt} = 0.32\).

1. Treating \(N\) and \(t\) as continuous variables, show that they satisfy the differential equation $$\frac{dN}{dt} = 0.02N(1 - 0.01N).$$ [1]
2. Solve the differential equation, obtaining an expression for \(t\) in terms of \(N\). [8]
3. Find the time at which the population will be double its value at \(t = 0\). [1]

In a certain chemical process a substance \(A\) reacts with and reduces a substance \(B\). The masses of \(A\) and \(B\) at time \(t\) after the start of the process are \(x\) and \(y\) respectively. It is given that \(\frac{dy}{dt} = -0.2xy\) and \(x = \frac{10}{(1 + t)^2}\). At the beginning of the process \(y = 100\).

1. Form a differential equation in \(y\) and \(t\), and solve this differential equation. [6]
2. Find the exact value approached by the mass of \(B\) as \(t\) becomes large. State what happens to the mass of \(A\) as \(t\) becomes large. [2]

\includegraphics{figure_10} A tank containing water is in the form of a cone with vertex \(C\). The axis is vertical and the semi-vertical angle is \(60°\), as shown in the diagram. At time \(t = 0\), the tank is full and the depth of water is \(H\). At this instant, a tap at \(C\) is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to \(\sqrt{h}\), where \(h\) is the depth of water at time \(t\). The tank becomes empty when \(t = 60\).

1. Show that \(h\) and \(t\) satisfy a differential equation of the form $$\frac{dh}{dt} = -Ah^{-\frac{1}{2}},$$ where \(A\) is a positive constant. [4]
2. Solve the differential equation given in part (i) and obtain an expression for \(t\) in terms of \(h\) and \(H\). [6]
3. Find the time at which the depth reaches \(\frac{1}{2}H\). [1]

[The volume \(V\) of a cone of vertical height \(h\) and base radius \(r\) is given by \(V = \frac{1}{3}\pi r^2 h\).]

The variables \(x\) and \(y\) satisfy the differential equation $$\frac{dy}{dx} = 4 \cos^2 y \tan x,$$ for \(0 \leqslant x < \frac{1}{2}\pi\), and \(x = 0\) when \(y = \frac{1}{4}\pi\). Solve this differential equation and find the value of \(x\) when \(y = \frac{1}{8}\pi\). [8]

A certain curve is such that its gradient at a general point with coordinates \((x, y)\) is proportional to \(\frac{y^2}{x}\). The curve passes through the points with coordinates \((1, 1)\) and \((e, 2)\). By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\). [8]

The coordinates \((x, y)\) of a general point on a curve satisfy the differential equation $$x\frac{dy}{dx} = (2 - x^2)y.$$ The curve passes through the point \((1, 1)\). Find the equation of the curve, obtaining an expression for \(y\) in terms of \(x\). [7]

A particle \(P\) of mass \(0.4\) kg is released from rest at a point \(O\) on a smooth plane inclined at \(30°\) to the horizontal. When the displacement of \(P\) from \(O\) is \(x\) m down the plane, the velocity of \(P\) is \(v \text{ ms}^{-1}\). A force of magnitude \(0.8e^{-5x}\) N acts on \(P\) up the plane along the line of greatest slope through \(O\).

1. Show that \(v \frac{dv}{dx} = 5 - 2e^{-x}\). [2]
2. Find \(v\) when \(x = 0.6\). [4]

\includegraphics{figure_6} A particle \(P\) of mass \(0.2\) kg is projected with velocity \(2\) m s\(^{-1}\) upwards along a line of greatest slope on a plane inclined at \(30°\) to the horizontal (see diagram). Air resistance of magnitude \(0.5v\) N opposes the motion of \(P\), where \(v\) m s\(^{-1}\) is the velocity of \(P\) at time \(t\) s after projection. The coefficient of friction between \(P\) and the plane is \(\frac{1}{2\sqrt{3}}\). The particle \(P\) reaches a position of instantaneous rest when \(t = T\).

1. Show that, while \(P\) is moving up the plane, \(\frac{dv}{dt} = -2.5(3 + v)\). [3]
2. Calculate \(T\). [4]
3. Calculate the speed of \(P\) when \(t = 2T\). [5]

A ball of mass 0.05 kg is released from rest at a height \(h\) m above the ground. At time \(t\) s after its release, the downward velocity of the ball is \(v\) m s\(^{-1}\). Air resistance opposes the motion of the ball with a force of magnitude 0.01\(v\) N.

1. Show that \(\frac{dv}{dt} = 10 - 0.2v\). Hence find \(v\) in terms of \(t\). [6]
2. Given that the ball reaches the ground when \(t = 2\), calculate \(h\). [4]

A particle of mass \(m\) moves in a straight line. At time \(t\), its displacement from a fixed point on the line is \(s\) and its velocity is \(v\). The particle experiences a retarding force of magnitude \(mkv^2\), where \(k\) is a positive constant. Find the relationship between \(v\) and \(t\). [7]

A particle \(P\) of mass \(2\) kg moving on a horizontal straight line has displacement \(x\) m from a fixed point \(O\) on the line and velocity \(v\) m s\(^{-1}\) at time \(t\) s. The only horizontal force acting on \(P\) has magnitude \(\frac{1}{10}(2v - 1)^2 e^{-t}\) N and acts towards \(O\). When \(t = 0\), \(x = 1\) and \(v = 3\).

1. Find an expression for \(v\) in terms of \(t\). [5]
2. Find an expression for \(x\) in terms of \(t\). [4]

A particle \(P\) moving in a straight line has displacement \(x\) m from a fixed point \(O\) on the line at time \(t\) s. The acceleration of \(P\), in m s\(^{-2}\), is given by \(\frac{200}{x^2} - \frac{100}{x^3}\) for \(x > 0\). When \(t = 0\), \(x = 1\) and \(P\) has velocity \(10\) m s\(^{-1}\) directed towards \(O\).

1. Show that the velocity \(v\) m s\(^{-1}\) of \(P\) is given by \(v = \frac{10(1-2x)}{x}\). [5]
2. Show that \(x\) and \(t\) are related by the equation \(e^{-40t} = (2x-1)e^{2x-2}\) and deduce what happens to \(x\) as \(t\) becomes large. [5]