4.10a General/particular solutions: of differential equations

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.

9 A biologist is investigating the spread of a weed in a particular region. At time \(t\) weeks after the start of the investigation, the area covered by the weed is \(A \mathrm {~m} ^ { 2 }\). The biologist claims that the rate of increase of \(A\) is proportional to \(\sqrt { } ( 2 A - 5 )\).

1. Write down a differential equation representing the biologist's claim.
2. At the start of the investigation, the area covered by the weed was \(7 \mathrm {~m} ^ { 2 }\) and, 10 weeks later, the area covered was \(27 \mathrm {~m} ^ { 2 }\). Assuming that the biologist's claim is correct, find the area covered 20 weeks after the start of the investigation.

8 The variables \(x\) and \(\theta\) satisfy the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} \theta } = ( x + 2 ) \sin ^ { 2 } 2 \theta$$ and it is given that \(x = 0\) when \(\theta = 0\). Solve the differential equation and calculate the value of \(x\) when \(\theta = \frac { 1 } { 4 } \pi\), giving your answer correct to 3 significant figures.

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.

3 The variables \(t\) and \(x\) are related by the differential equation $$\frac { d ^ { 2 } x } { d t ^ { 2 } } + \frac { d x } { d t } + x = t ^ { 2 } + 1$$

1. Find the general solution for \(x\) in terms of \(t\).
2. Deduce an approximate value of \(\frac { \mathrm { d } ^ { 2 } \mathrm { x } } { \mathrm { dt } ^ { 2 } }\) for large positive values of \(t\).

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}

7 The variables \(x\) and \(y\) are related by the differential equation $$4 \frac { d ^ { 2 } y } { d x ^ { 2 } } - y = 3$$ It is given that, when \(x = 0 , y = - 3\) and \(\frac { \mathrm { dy } } { \mathrm { dx } } = 2\).

1. Find \(y\) in terms of \(x\).
2. Deduce the exact value of \(x\) for which \(y = 0\). Give your answer in logarithmic form.

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.

1. In freezing temperatures, ice forms on the surface of the water in a barrel. At time \(t\) hours after the start of freezing, the thickness of the ice formed is \(x \mathrm {~mm}\). You may assume that the thickness of the ice is uniform across the surface of the water.

At 4 pm there is no ice on the surface, and freezing begins.
At 6pm, after two hours of freezing, the ice is 1.5 mm thick.
In a simple model, the rate of increase of \(x\), in mm per hour, is assumed to be constant for a period of 20 hours. Using this simple model,

1. express \(t\) in terms of \(x\),
2. find the value of \(t\) when \(x = 3\) In a second model, the rate of increase of \(x\), in mm per hour, is given by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { \lambda } { ( 2 x + 1 ) } \text { where } \lambda \text { is a constant and } 0 \leqslant t \leqslant 20$$ Using this second model,
3. solve the differential equation and express \(t\) in terms of \(x\) and \(\lambda\),
4. find the exact value for \(\lambda\),
5. find at what time the ice is predicted to be 3 mm thick.

4. Given that $$y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 y = 0$$

1. show that $$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \frac { 28 } { y ^ { 2 } } \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 3 } - \frac { 24 } { y } \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right)$$ Given also that \(y = 8\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) at \(x = 0\)
2. find a series solution for \(y\) in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\), simplifying the coefficients where possible.

1. Given that

$$y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } - 2 y = 0 \quad y > 0$$

1. determine \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\) in terms of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } , \frac { \mathrm {~d} y } { \mathrm {~d} x }\) and \(y\) Given that \(y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) at \(x = 0\)
2. determine a series solution for \(y\) in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\), giving each coefficient in its simplest form.

2. $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } - y ^ { 3 } = 4$$

1. Show that $$x \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = a y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + \left( b y ^ { 2 } + c \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$$ where \(a\), \(b\) and \(c\) are integers to be determined. Given that \(y = 1\) at \(x = 2\)
2. determine the Taylor series expansion for \(y\) in ascending powers of \(( x - 2 )\), up to and including the term in \(( x - 2 ) ^ { 3 }\), giving each coefficient in simplest form.

1. (a) Show that the substitution \(y ^ { 2 } = \frac { 1 } { t }\) transforms the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + y = x y ^ { 3 }$$ into the differential equation $$\frac { \mathrm { d } t } { \mathrm {~d} x } - 2 t = - 2 x$$ (b) Solve differential equation (II) and determine \(y ^ { 2 }\) in terms of \(x\).

12. (a) Use the substitution \(y = v x\) to transform the equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 4 x + y ) ( x + y ) } { x ^ { 2 } } , x > 0$$ into the equation $$x \frac { \mathrm {~d} v } { \mathrm {~d} x } = ( 2 + v ) ^ { 2 }$$ (b) Solve the differential equation II to find \(\boldsymbol { v }\) as a function of \(\boldsymbol { x }\) (c) Hence show that \(\quad y = - 2 x - \frac { x } { \ln x + c }\), where \(c\) is an arbitrary constant, is a general solution of the differential equation I.

3. (a) Show that the transformation \(y = x v\) transforms the equation $$\begin{array} { l l } x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + \left( 2 + 9 x ^ { 2 } \right) y = x ^ { 5 } , \\ \text { into the equation } & \frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 9 v = x ^ { 2 } . \end{array}$$I (b) Solve the differential equation II to find \(v\) as a function of \(x\).
(c) Hence state the general solution of the differential equation I.
(1)(Total 12 marks)

7. (a) Find the general solution of the differential equation $$2 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 x = 2 t + 9$$ (b) Find the particular solution of this differential equation for which \(x = 3\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - 1\) when \(t = 0\). The particular solution in part (b) is used to model the motion of a particle \(P\) on the \(x\)-axis. At time \(t\) seconds \(( t \geq 0 ) , P\) is \(x\) metres from the origin \(O\).
(c) Show that the minimum distance between \(O\) and \(P\) is \(\frac { 1 } { 2 } ( 5 + \ln 2 ) \mathrm { m }\) and justify that the distance is a minimum.
(4)(Total 14 marks)

3. A scientist is modelling the amount of a chemical in the human bloodstream. The amount \(x\) of the chemical, measured in \(\mathrm { mg } l ^ { - 1 }\), at time \(t\) hours satisfies the differential equation $$2 x \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 6 \left( \frac { \mathrm { dx } } { \mathrm { dt } } \right) ^ { 2 } = x ^ { 2 } - 3 x ^ { 4 } , \quad x > 0$$

1. Show that the substitution \(\mathrm { y } = \frac { 1 } { x ^ { 2 } }\) transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + y = 3$$
2. Find the general solution of differential equation \(I\). Given that at time \(t = 0 , x = \frac { 1 } { 2 }\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\),
3. find an expression for \(x\) in terms of \(t\),
4. write down the maximum value of \(x\) as \(t\) varies.

9. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y \mathrm { e } ^ { x ^ { 2 } } .$$ It is given that \(y = 0.2\) at \(x = 0\).

1. Use the approximation \(\frac { y _ { 1 } - y _ { 0 } } { h } \approx \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) _ { 0 }\), with \(h = 0.1\), to obtain an estimate of the value of \(y\) at \(x = 0.1\).
2. Use your answer to part (a) and the approximation \(\frac { y _ { 2 } - y _ { 0 } } { 2 h } \approx \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) _ { 1 }\), with \(h = 0.1\), to obtain an estimate of the value of \(y\) at \(x = 0.2\). Gives your answer to 4 decimal places.
   (Total 5 marks)

2. The displacement \(x\) metres of a particle at time \(t\) seconds is given by the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + x + \cos x = 0$$ When \(t = 0 , x = 0\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 } { 2 }\).
Find a Taylor series solution for \(x\) in ascending powers of \(t\), up to and including the term in \(t ^ { 3 }\).

4. (i) $$p \frac { \mathrm {~d} x } { \mathrm {~d} t } + q x = r \quad \text { where } p , q \text { and } r \text { are constants }$$ Given that \(x = 0\) when \(t = 0\)

1. find \(x\) in terms of \(t\)
2. find the limiting value of \(x\) as \(t \rightarrow \infty\) (ii) $$\frac { \mathrm { d } y } { \mathrm {~d} \theta } + 2 y = \sin \theta$$ Given that \(y = 0\) when \(\theta = 0\), find \(y\) in terms of \(\theta\)

4. (a) Determine the general solution of the differential equation $$( x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } - x y = \mathrm { e } ^ { 3 x } \quad x > - 1$$ giving your answer in the form \(y = \mathrm { f } ( x )\).
(b) Determine the particular solution of the differential equation for which \(y = 5\) when \(x = 0\)

7. (a) Show that the transformation \(x = t ^ { 2 }\) transforms the differential equation $$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 15 y = 15 x$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } - 15 y = 15 t ^ { 2 }$$ (b) Solve differential equation (II) to determine \(y\) in terms of \(t\).
(c) Hence determine the general solution of differential equation (I). \includegraphics[max width=\textwidth, alt={}, center]{8fa1e7da-009f-4b7f-9fa8-21a1768bfd73-24_2258_53_308_1980}

7 A quantity of oil is dropped into water. When the oil hits the water it spreads out as a circle. The radius of the circle is \(r \mathrm {~cm}\) after \(t\) seconds and when \(t = 3\) the radius of the circle is increasing at the rate of 0.5 centimetres per second.
One observer believes that the radius increases at a rate which is proportional to \(\frac { 1 } { ( t + 1 ) }\).

1. Write down a differential equation for this situation, using \(k\) as a constant of proportionality.
2. Show that \(k = 2\).
3. Calculate the radius of the circle after 10 seconds according to this model. Another observer believes that the rate of increase of the radius of the circle is proportional to \(\frac { 1 } { ( t + 1 ) ( t + 2 ) }\).
4. Write down a new differential equation for this new situation. Using the same initial conditions as before, find the value of the new constant of proportionality.
5. Hence solve the differential equation.
6. Calculate the radius of the circle after 10 seconds according to this model.

5 The variables \(x\) and \(y\) are related by the differential equation $$x ^ { 3 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = x y + x + 1 .$$

1. Use the substitution \(y = u - \frac { 1 } { x }\), where \(u\) is a function of \(x\), to show that the differential equation may be written as $$x ^ { 2 } \frac { \mathrm {~d} u } { \mathrm {~d} x } = u .$$
2. Hence find the general solution of the differential equation (A), giving your answer in the form \(y = \mathrm { f } ( x )\).