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

9

1. Express \(\frac { 1 } { x ( 2 x + 3 ) }\) in partial fractions.
2. The variables \(x\) and \(y\) satisfy the differential equation $$x ( 2 x + 3 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = y$$ and it is given that \(y = 1\) when \(x = 1\). Solve the differential equation and calculate the value of \(y\) when \(x = 9\), giving your answer correct to 3 significant figures.

8 In a certain chemical reaction, a compound \(A\) is formed from a compound \(B\). The masses of \(A\) and \(B\) at time \(t\) after the start of the reaction are \(x\) and \(y\) respectively and the sum of the masses is equal to 50 throughout the reaction. At any time the rate of increase of the mass of \(A\) is proportional to the mass of \(B\) at that time.

1. Explain why \(\frac { \mathrm { d } x } { \mathrm {~d} t } = k ( 50 - x )\), where \(k\) is a constant.
   It is given that \(x = 0\) when \(t = 0\), and \(x = 25\) when \(t = 10\).
2. Solve the differential equation in part (i) and express \(x\) in terms of \(t\).

5

1. Differentiate \(\frac { 1 } { \sin ^ { 2 } \theta }\) with respect to \(\theta\).
2. The variables \(x\) and \(\theta\) satisfy the differential equation $$x \tan \theta \frac { d x } { d \theta } + \operatorname { cosec } ^ { 2 } \theta = 0$$ for \(0 < \theta < \frac { 1 } { 2 } \pi\) and \(x > 0\). It is given that \(x = 4\) when \(\theta = \frac { 1 } { 6 } \pi\). Solve the differential equation, obtaining an expression for \(x\) in terms of \(\theta\).

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

1. Solve the differential equation, obtaining \(y\) in terms of \(x\).
2. Explain why \(x\) can only take values that are less than 1 .

5 The variables \(x\) and \(y\) satisfy the differential equation $$( x + 1 ) y \frac { \mathrm {~d} y } { \mathrm {~d} x } = y ^ { 2 } + 5$$ It is given that \(y = 2\) when \(x = 0\). Solve the differential equation obtaining an expression for \(y ^ { 2 }\) in terms of \(x\).

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

1. Solve the differential equation and obtain an expression for \(y\) in terms of \(x\).
2. Explain briefly why \(x\) can only take values less than 1 .

7 \includegraphics[max width=\textwidth, alt={}, center]{e26f21c5-3776-4c86-8440-6959c5e37486-12_444_382_258_886} A water tank has vertical sides and a horizontal rectangular base, as shown in the diagram. The area of the base is \(2 \mathrm {~m} ^ { 2 }\). At time \(t = 0\) the tank is empty and water begins to flow into it at a rate of \(1 \mathrm {~m} ^ { 3 }\) per hour. At the same time water begins to flow out from the base at a rate of \(0.2 \sqrt { } h \mathrm {~m} ^ { 3 }\) per hour, where \(h \mathrm {~m}\) is the depth of water in the tank at time \(t\) hours.

1. Form a differential equation satisfied by \(h\) and \(t\), and show that the time \(T\) hours taken for the depth of water to reach 4 m is given by $$T = \int _ { 0 } ^ { 4 } \frac { 10 } { 5 - \sqrt { } h } \mathrm {~d} h$$
2. Using the substitution \(u = 5 - \sqrt { } h\), find the value of \(T\).

6 The variables \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = k y ^ { 3 } \mathrm { e } ^ { - x }$$ where \(k\) is a constant. It is given that \(y = 1\) when \(x = 0\), and that \(y = \sqrt { } \mathrm { e }\) when \(x = 1\). Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).

9 In an experiment to study the spread of a soil disease, an area of \(10 \mathrm {~m} ^ { 2 }\) of soil was exposed to infection. In a simple model, it is assumed that the infected area grows at a rate which is proportional to the product of the infected area and the uninfected area. Initially, \(5 \mathrm {~m} ^ { 2 }\) was infected and the rate of growth of the infected area was \(0.1 \mathrm {~m} ^ { 2 }\) per day. At time \(t\) days after the start of the experiment, an area \(a \mathrm {~m} ^ { 2 }\) is infected and an area \(( 10 - a ) \mathrm { m } ^ { 2 }\) is uninfected.

1. Show that \(\frac { \mathrm { d } a } { \mathrm {~d} t } = 0.004 a ( 10 - a )\).
2. By first expressing \(\frac { 1 } { a ( 10 - a ) }\) in partial fractions, solve this differential equation, obtaining an expression for \(t\) in terms of \(a\).
3. Find the time taken for \(90 \%\) of the soil area to become infected, according to this model.

9 Compressed air is escaping from a container. The pressure of the air in the container at time \(t\) is \(P\), and the constant atmospheric pressure of the air outside the container is \(A\). The rate of decrease of \(P\) is proportional to the square root of the pressure difference ( \(P - A\) ). Thus the differential equation connecting \(P\) and \(t\) is $$\frac { \mathrm { d } P } { \mathrm {~d} t } = - k \sqrt { } ( P - A )$$ where \(k\) is a positive constant.

1. Find, in any form, the general solution of this differential equation.
2. Given that \(P = 5 A\) when \(t = 0\), and that \(P = 2 A\) when \(t = 2\), show that \(k = \sqrt { } A\).
3. Find the value of \(t\) when \(P = A\).
4. Obtain an expression for \(P\) in terms of \(A\) and \(t\).

10 A rectangular reservoir has a horizontal base of area \(1000 \mathrm {~m} ^ { 2 }\). At time \(t = 0\), it is empty and water begins to flow into it at a constant rate of \(30 \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }\). At the same time, water begins to flow out at a rate proportional to \(\sqrt { } h\), where \(h \mathrm {~m}\) is the depth of the water at time \(t \mathrm {~s}\). When \(h = 1 , \frac { \mathrm {~d} h } { \mathrm {~d} t } = 0.02\).

1. Show that \(h\) satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.01 ( 3 - \sqrt { } h )$$ It is given that, after making the substitution \(x = 3 - \sqrt { } h\), the equation in part (i) becomes $$( x - 3 ) \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.005 x$$
2. Using the fact that \(x = 3\) when \(t = 0\), solve this differential equation, obtaining an expression for \(t\) in terms of \(x\).
3. Find the time at which the depth of water reaches 4 m .

8 In a certain chemical reaction the amount, \(x\) grams, of a substance present is decreasing. The rate of decrease of \(x\) is proportional to the product of \(x\) and the time, \(t\) seconds, since the start of the reaction. Thus \(x\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - k x t$$ where \(k\) is a positive constant. At the start of the reaction, when \(t = 0 , x = 100\).

1. Solve this differential equation, obtaining a relation between \(x , k\) and \(t\).
2. 20 seconds after the start of the reaction the amount of substance present is 90 grams. Find the time after the start of the reaction at which the amount of substance present is 50 grams.

4 Given that \(y = 2\) when \(x = 0\), solve the differential equation $$y \frac { \mathrm {~d} y } { \mathrm {~d} x } = 1 + y ^ { 2 }$$ obtaining an expression for \(y ^ { 2 }\) in terms of \(x\).

6

1. By sketching a suitable pair of graphs, show that the equation $$2 - x = \ln x$$ has only one root.
2. Verify by calculation that this root lies between 1.4 and 1.7.
3. Show that this root also satisfies the equation $$x = \frac { 1 } { 3 } ( 4 + x - 2 \ln x )$$
4. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 3 } \left( 4 + x _ { n } - 2 \ln x _ { n } \right)$$ with initial value \(x _ { 1 } = 1.5\), to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

8 \includegraphics[max width=\textwidth, alt={}, center]{c687888e-bef0-4ea9-b5b3-e614028cc07c-3_654_805_274_671} An underground storage tank is being filled with liquid as shown in the diagram. Initially the tank is empty. At time \(t\) hours after filling begins, the volume of liquid is \(V \mathrm {~m} ^ { 3 }\) and the depth of liquid is \(h \mathrm {~m}\). It is given that \(V = \frac { 4 } { 3 } h ^ { 3 }\). The liquid is poured in at a rate of \(20 \mathrm {~m} ^ { 3 }\) per hour, but owing to leakage, liquid is lost at a rate proportional to \(h ^ { 2 }\). When \(h = 1 , \frac { \mathrm {~d} h } { \mathrm {~d} t } = 4.95\).

1. Show that \(h\) satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { 5 } { h ^ { 2 } } - \frac { 1 } { 20 } .$$
2. Verify that \(\frac { 20 h ^ { 2 } } { 100 - h ^ { 2 } } \equiv - 20 + \frac { 2000 } { ( 10 - h ) ( 10 + h ) }\).
3. Hence solve the differential equation in part (i), obtaining an expression for \(t\) in terms of \(h\).

10 In a model of the expansion of a sphere of radius \(r \mathrm {~cm}\), it is assumed that, at time \(t\) seconds after the start, the rate of increase of the surface area of the sphere is proportional to its volume. When \(t = 0\), \(r = 5\) and \(\frac { \mathrm { d } r } { \mathrm {~d} t } = 2\).

1. Show that \(r\) satisfies the differential equation $$\frac { \mathrm { d } r } { \mathrm {~d} t } = 0.08 r ^ { 2 }$$ [The surface area \(A\) and volume \(V\) of a sphere of radius \(r\) are given by the formulae \(A = 4 \pi r ^ { 2 }\), \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\).]
2. Solve this differential equation, obtaining an expression for \(r\) in terms of \(t\).
3. Deduce from your answer to part (ii) the set of values that \(t\) can take, according to this model.

10 A certain substance is formed in a chemical reaction. The mass of substance formed \(t\) seconds after the start of the reaction is \(x\) grams. At any time the rate of formation of the substance is proportional to \(( 20 - x )\). When \(t = 0 , x = 0\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 1\).

1. Show that \(x\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.05 ( 20 - x ) .$$
2. Find, in any form, the solution of this differential equation.
3. Find \(x\) when \(t = 10\), giving your answer correct to 1 decimal place.
4. State what happens to the value of \(x\) as \(t\) becomes very large.

4 The variables \(x\) and \(\theta\) are related by the differential equation $$\sin 2 \theta \frac { \mathrm {~d} x } { \mathrm {~d} \theta } = ( x + 1 ) \cos 2 \theta$$ where \(0 < \theta < \frac { 1 } { 2 } \pi\). When \(\theta = \frac { 1 } { 12 } \pi , x = 0\). Solve the differential equation, obtaining an expression for \(x\) in terms of \(\theta\), and simplifying your answer as far as possible.

4 During an experiment, the number of organisms present at time \(t\) days is denoted by \(N\), where \(N\) is treated as a continuous variable. It is given that $$\frac { \mathrm { d } N } { \mathrm {~d} t } = 1.2 \mathrm { e } ^ { - 0.02 t } N ^ { 0.5 }$$ When \(t = 0\), the number of organisms present is 100 .

1. Find an expression for \(N\) in terms of \(t\).
2. State what happens to the number of organisms present after a long time.

6 The variables \(x\) and \(y\) are related by the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 1 - y ^ { 2 }$$ When \(x = 2 , y = 0\). Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).

4 The variables \(x\) and \(y\) are related by the differential equation $$\left( x ^ { 2 } + 4 \right) \frac { d y } { d x } = 6 x y$$ It is given that \(y = 32\) when \(x = 0\). Find an expression for \(y\) in terms of \(x\).

10 \includegraphics[max width=\textwidth, alt={}, center]{dd7b2aee-4318-48e8-97c0-541e47f2e83a-4_335_875_262_635} A tank containing water is in the form of a cone with vertex \(C\). The axis is vertical and the semivertical angle is \(60 ^ { \circ }\), 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 $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - A h ^ { - \frac { 3 } { 2 } } ,$$ where \(A\) is a positive constant.
2. Solve the differential equation given in part (i) and obtain an expression for \(t\) in terms of \(h\) and \(H\).
3. Find the time at which the depth reaches \(\frac { 1 } { 2 } H\).
   [0pt] [The volume \(V\) of a cone of vertical height \(h\) and base radius \(r\) is given by \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\).]

10 \includegraphics[max width=\textwidth, alt={}, center]{3621a7e5-a3fb-42c1-828d-7068fddbf2f9-3_677_691_781_724} A particular solution of the differential equation $$3 y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = 4 \left( y ^ { 3 } + 1 \right) \cos ^ { 2 } x$$ is such that \(y = 2\) when \(x = 0\). The diagram shows a sketch of the graph of this solution for \(0 \leqslant x \leqslant 2 \pi\); the graph has stationary points at \(A\) and \(B\). Find the \(y\)-coordinates of \(A\) and \(B\), giving each coordinate correct to 1 decimal place.

7 In a certain country the government charges tax on each litre of petrol sold to motorists. The revenue per year is \(R\) million dollars when the rate of tax is \(x\) dollars per litre. The variation of \(R\) with \(x\) is modelled by the differential equation $$\frac { \mathrm { d } R } { \mathrm {~d} x } = R \left( \frac { 1 } { x } - 0.57 \right)$$ where \(R\) and \(x\) are taken to be continuous variables. When \(x = 0.5 , R = 16.8\).

1. Solve the differential equation and obtain an expression for \(R\) in terms of \(x\).
2. This model predicts that \(R\) cannot exceed a certain amount. Find this maximum value of \(R\).

8 The variables \(x\) and \(y\) are related by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 5 } x y ^ { \frac { 1 } { 2 } } \sin \left( \frac { 1 } { 3 } x \right)$$

1. Find the general solution, giving \(y\) in terms of \(x\).
2. Given that \(y = 100\) when \(x = 0\), find the value of \(y\) when \(x = 25\).