4.10c Integrating factor: first order equations

9 The temperature of a quantity of liquid at time \(t\) is \(\theta\). The liquid is cooling in an atmosphere whose temperature is constant and equal to \(A\). The rate of decrease of \(\theta\) is proportional to the temperature difference \(( \theta - A )\). Thus \(\theta\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta - A )$$ where \(k\) is a positive constant.

1. Find, in any form, the solution of this differential equation, given that \(\theta = 4 A\) when \(t = 0\).
2. Given also that \(\theta = 3 A\) when \(t = 1\), show that \(k = \ln \frac { 3 } { 2 }\).
3. Find \(\theta\) in terms of \(A\) when \(t = 2\), expressing your answer in its simplest form.

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.

10 A large plantation of area \(20 \mathrm {~km} ^ { 2 }\) is becoming infected with a plant disease. At time \(t\) years the area infected is \(x \mathrm {~km} ^ { 2 }\) and the rate of increase of \(x\) is proportional to the ratio of the area infected to the area not yet infected. When \(t = 0 , x = 1\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 1\).

1. Show that \(x\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 19 x } { 20 - x }$$
2. Solve the differential equation and show that when \(t = 1\) the value of \(x\) satisfies the equation \(x = \mathrm { e } ^ { 0.9 + 0.05 x }\).
3. Use an iterative formula based on the equation in part (b), with an initial value of 2 , to determine \(x\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
4. Calculate the value of \(t\) at which the entire plantation becomes infected.

1 Find the solution of the differential equation $$\frac { d y } { d x } + 5 y = e ^ { - 7 x }$$ for which \(y = 0\) when \(x = 0\). Give your answer in the form \(y = f ( x )\).

4 Find the solution of the differential equation $$\sin \theta \frac { d y } { d \theta } + y = \tan \frac { 1 } { 2 } \theta$$ where \(0 < \theta < \pi\), given that \(y = 1\) when \(\theta = \frac { 1 } { 2 } \pi\). Give your answer in the form \(y = \mathrm { f } ( \theta )\). [You may use without proof the result that \(\int \operatorname { cosec } \theta d \theta = \ln \tan \frac { 1 } { 2 } \theta\).]

6

1. Starting from the definitions of sinh and cosh in terms of exponentials, prove that $$2 \sinh ^ { 2 } x = \cosh 2 x - 1$$ \includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_67_1550_374_347} \includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_65_1569_468_328} \includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_67_1573_557_324} \includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_70_1573_646_324} \includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_72_1573_735_324} \includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_72_1570_826_324} \includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_74_1570_916_324} \includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_69_1570_1007_324}
2. Find the solution to the differential equation $$\frac { d y } { d x } + y \operatorname { coth } x = 4 \sinh x$$ for which \(y = 1\) when \(x = \ln 3\).

5 Find the solution of the differential equation $$x ( x + 7 ) \frac { d y } { d x } + 7 y = x$$ for which \(y = 7\) when \(x = 1\). Give your answer in the form \(y = f ( x )\).

5

1. Starting from the definitions of cosh and sinh in terms of exponentials, prove that $$2 \cosh ^ { 2 } x = \cosh 2 x + 1$$ \includegraphics[max width=\textwidth, alt={}, center]{d421652f-576d-4843-abbf-54404e225fec-08_67_1550_374_347}
2. Find the solution of the differential equation $$\frac { d y } { d x } + 2 y \tanh x = 1$$ for which \(y = 1\) when \(x = 0\). Give your answer in the form \(y = f ( x )\).

7

1. Use the substitution \(\mathrm { u } = 1 + \mathrm { x } ^ { 2 }\) to find $$\int \frac { x } { \sqrt { 1 + x ^ { 2 } } } d x$$
2. Find the solution of the differential equation $$x \frac { d y } { d x } - y = x ^ { 2 } \sinh ^ { - 1 } x$$ given that \(y = 1\) when \(x = 1\). Give your answer in the form \(\mathrm { y } = \mathrm { f } ( \mathrm { x } )\).

7

1. Show that $$\frac { \mathrm { d } } { \mathrm {~d} x } \left( \frac { x } { 2 } \sqrt { x ^ { 2 } - 9 } - \frac { 9 } { 2 } \cosh ^ { - 1 } \frac { x } { 3 } \right) = \sqrt { x ^ { 2 } - 9 }$$ \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-14_67_1579_413_324} \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-14_77_1581_497_322}
2. Find the solution of the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } - y = x ^ { 2 } \sqrt { x ^ { 2 } - 9 }$$ given that \(y = 1\) when \(x = 3\). Give your answer in the form \(y = \mathrm { f } ( x )\). \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-14_2716_35_143_2012}

6

1. Use de Moivre's theorem to show that \(\sin ^ { 4 } \theta = \frac { 1 } { 8 } ( \cos 4 \theta - 4 \cos 2 \theta + 3 )\).
2. Find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} \theta } + y \cot \theta = \sin ^ { 3 } \theta$$ for which \(y = 0\) when \(\theta = \frac { 1 } { 2 } \pi\).

4 Find the solution of the differential equation $$x \frac { d y } { d x } + 2 y = e ^ { x }$$ for which \(y = 3\) when \(x = 1\). Give your answer in the form \(y = f ( x )\).

7

1. Show that an appropriate integrating factor for $$\sqrt { x ^ { 2 } - 1 } \frac { d y } { d x } + y = x ^ { 2 } - x \sqrt { x ^ { 2 } - 1 }$$ is \(x + \sqrt { x ^ { 2 } - 1 }\).
2. Hence find the solution of the differential equation $$\sqrt { x ^ { 2 } - 1 } \frac { d y } { d x } + y = x ^ { 2 } - x \sqrt { x ^ { 2 } - 1 }$$ for which \(y = 1\) when \(x = \frac { 5 } { 4 }\). Give your answer in the form \(y = f ( x )\).

8

1. Use the substitution \(u = 1 - ( \theta - 1 ) ^ { 2 }\) to find $$\int \frac { \theta - 1 } { \sqrt { 1 - ( \theta - 1 ) ^ { 2 } } } \mathrm {~d} \theta$$
2. Find the solution of the differential equation $$\theta \frac { d y } { d \theta } - y = \theta ^ { 2 } \sin ^ { - 1 } ( \theta - 1 ) ,$$ where \(0 < \theta < 2\), given that \(y = 1\) when \(\theta = 1\). Give your answer in the form \(y = \mathrm { f } ( \theta )\).
   If you use the following page to complete the answer to any question, the question number must be clearly shown.

4 Find the solution of the differential equation $$\left( 4 t ^ { 2 } - 1 \right) \frac { d x } { d t } + 4 x = 4 t ^ { 2 } - 1$$ for which \(x = 3\) when \(t = 1\). Give your answer in the form \(\mathrm { x } = \mathrm { f } ( \mathrm { t } )\).

6

1. Starting from the definitions of cosh and sinh in terms of exponentials, prove that $$\sinh 2 x = 2 \sinh x \cosh x$$ \includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_67_1550_374_347} \includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_58_1569_475_328} \includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_58_1569_566_328} \includegraphics[max width=\textwidth, alt={}]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_54_1566_657_328} ....................................................................................................................................................... . \includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_54_1570_840_324} \includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_53_1570_932_324} \includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_53_1570_1023_324}
2. Using the substitution \(\mathrm { u } = \sinh \mathrm { x }\), find \(\int \sinh ^ { 2 } 2 x \cosh x \mathrm { dx }\).
3. Find the particular solution of the differential equation $$\frac { d y } { d x } + y \tanh x = \sinh ^ { 2 } 2 x$$ given that \(y = 4\) when \(x = 0\). Give your answer in the form \(y = f ( x )\).

4 Find the solution of the differential equation $$\frac { d y } { d x } + 3 y = \sin x$$ for which \(y = 1\) when \(x = 0\). Give your answer in the form \(y = f ( x )\).

3 Find the solution of the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = \frac { \sin x } { x }$$ for which \(y = 0\) when \(x = \frac { 1 } { 2 } \pi\). Give your answer in the form \(y = \mathrm { f } ( x )\).

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 \(t\).

14. The volume of a spherical balloon of radius \(r \mathrm {~cm}\) is \(V \mathrm {~cm} ^ { 3 }\), where \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\)

1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\) The volume of the balloon increases with time \(t\) seconds according to the formula $$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 9000 \pi } { ( t + 81 ) ^ { \frac { 5 } { 4 } } } \quad t \geqslant 0$$
2. Using the chain rule, or otherwise, show that $$\frac { \mathrm { d } r } { \mathrm {~d} t } = \frac { k } { r ^ { n } ( t + 81 ) ^ { \frac { 5 } { 4 } } } \quad t \geqslant 0$$ where \(k\) and \(n\) are constants to be found. Initially, the radius of the balloon is 3 cm .
3. Using the values of \(k\) and \(n\) found in part (b), solve the differential equation $$\frac { \mathrm { d } r } { \mathrm {~d} t } = \frac { k } { r ^ { n } ( t + 81 ) ^ { \frac { 5 } { 4 } } } \quad t \geqslant 0$$ to obtain a formula for \(r\) in terms of \(t\).
4. Hence find the radius of the balloon when \(t = 175\), giving your answer to 3 significant figures.
   (1)
5. Find the rate of increase of the radius of the balloon when \(t = 175\). Give your answer to 3 significant figures.

   END

4. (a) Show that the substitution \(y ^ { 2 } = \frac { 1 } { z }\) transforms the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y = 3 x y ^ { 3 } \quad y \neq 0$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - 4 z = - 6 x$$ (b) Obtain the general solution of differential equation (II).
(c) Hence obtain the general solution of differential equation (I), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\)

1. (a) Show that the transformation \(v = y - 2 x\) transforms the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y x ( y - 4 x ) = 2 - 8 x ^ { 3 }$$ into the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} x } = - 2 x v ^ { 2 }$$ (b) Solve the differential equation (II) to determine \(v\) as a function of \(x\) (c) Hence obtain the general solution of the differential equation (I).
(d) Sketch the solution curve that passes through the point \(( - 1 , - 1 )\). On your sketch show clearly the equation of any horizontal or vertical asymptotes.
You do not need to find the coordinates of any intercepts with the coordinate axes or the coordinates of any stationary points.

1. (a) Show that the transformation \(y = \frac { 1 } { z }\) transforms the differential equation

$$x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + x y = 2 y ^ { 2 }$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - \frac { z } { x } = - \frac { 2 } { x ^ { 2 } }$$ (b) Solve differential equation (II) to determine \(z\) in terms of \(x\).
(c) Hence determine the particular solution of differential equation (I) for which \(y = - \frac { 3 } { 8 }\) at \(x = 3\) Give your answer in the form \(y = \mathrm { f } ( x )\).

1. (a) For all the values of \(x\) where the identity is defined, prove that

$$\cot 2 x + \tan x \equiv \operatorname { cosec } 2 x$$ (b) Show that the substitution \(y ^ { 2 } = w \sin 2 x\), where \(w\) is a function of \(x\), transforms the differential equation $$y \frac { \mathrm {~d} y } { \mathrm {~d} x } + y ^ { 2 } \tan x = \sin x \quad 0 < x < \frac { \pi } { 2 }$$ into the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} x } + 2 w \operatorname { cosec } 2 x = \sec x \quad 0 < x < \frac { \pi } { 2 }$$ (c) By solving differential equation (II), determine a general solution of differential equation (I) in the form \(y ^ { 2 } = \mathrm { f } ( x )\), where \(\mathrm { f } ( x )\) is a function in terms of \(\cos x\) $$\text { [You may use without proof } \left. \int \operatorname { cosec } 2 x \mathrm {~d} x = \frac { 1 } { 2 } \ln | \tan x | \text { (+ constant) } \right]$$

6. $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + ( 1 - 2 x ) y = x , \quad x > 0$$ Find the general solution of the differential equation, giving your answer in the form \(y = \mathrm { f } ( x )\).