A-Level Maths (Edexcel 9MA0) · Pure Year 2 · Chapter 11: Integration · Lesson 8

Solving differential equations

112 marks

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Quick questions to start.

A1
Needed today
Find $\int \frac{1}{2 y + 1} d y$ .
[2 marks]
Answer
A2
Common slip
$ln ∣ y ∣ = 3 x + c$ . Write $y$ in terms of $x$ using a single constant $A$ .
[2 marks]
Answer
A3
From chapter 1
Express $\frac{3}{( x - 1 ) ( x + 2 )}$ in partial fractions.
[2 marks]
Answer
A4
Needed today
Find $\int x e^{x} d x$ .
[2 marks]
Answer

Example, then your turn

A-Level

Your teacher works through each example with you. Then try the one beside it.

Example 1

Given that $y = 2$ when $x = 0$ , solve $\frac{d y}{d x} = 6 x^{2} y$ , $y > 0$ , giving $y$ in terms of $x$ .

Hint Move every

y

to the left with

d y

and every

x

to the right with

d x

. Put one constant on the right only.

B1
Given that $y = 3$ when $x = 0$ , solve $\frac{d y}{d x} = y sin x$ , $y > 0$ , giving $y$ in terms of $x$ .
[4 marks]
Answer

Example 2

Given that $y = 2$ when $x = 0$ , solve $\frac{d y}{d x} = \frac{2 x + 1}{3 y ^{2}}$ , giving $y$ in terms of $x$ .

Hint Separate to

\int 3 y^{2} d y = \int (2 x + 1) d x

. Find the constant first, then make

y

the subject.

B2
Given that $y = 2$ when $x = 0$ , solve $\frac{d y}{d x} = \frac{x e ^{x}}{y}$ , $y > 0$ . Give $y^{2}$ in terms of $x$ .
[5 marks]
Answer

Example 3

Given that $y = 4$ when $x = 1$ , solve $\frac{d y}{d x} = \frac{2 y}{( x + 1 ) ( x + 3 )}$ , $x > - 1$ , $y > 0$ , giving $y$ in terms of $x$ .

Hint Write

\frac{2}{( x + 1 ) ( x + 3 )} = \frac{1}{x + 1} - \frac{1}{x + 3}

before integrating. Then combine the logs.

B3
Given that $y = 1$ when $x = 2$ , solve $\frac{d y}{d x} = \frac{3 y}{( x - 1 ) ( x + 2 )}$ , $x > 1$ , $y > 0$ , giving $y$ in terms of $x$ .
[5 marks]
Answer

Example 4

Given that $y = \frac{π}{4}$ when $x = 0$ , solve $\frac{d y}{d x} = e^{x} cos^{2} y$ , $- \frac{π}{2} < y < \frac{π}{2}$ .

Hint

\frac{1}{cos ^{2} y} = sec^{2} y

, and

\int sec^{2} y d y = tan y

B4
Given that $y = \frac{π}{6}$ when $x = \frac{π}{3}$ , solve $\frac{d y}{d x} = \frac{sin x}{cos y}$ , $- \frac{π}{2} < y < \frac{π}{2}$ , giving $sin y$ in terms of $x$ . Hence find $y$ when $x = \frac{π}{4}$ , to $3$ significant figures.
[5 marks]
Answer

Practice

A-Level

C1
Sam solves $\frac{d y}{d x} = 2 x y$ , $y > 0$ , given that $y = 5$ when $x = 0$ . He writes: " $ln y = x^{2}$ , so $y = e^{x^{2}} + c$ , and then $c = 4$ ." Explain his mistake. Find the correct particular solution, and the value of $y$ at $x = 1$ from his answer and from the correct one, to $3$ significant figures.
[5 marks]
Answer

C2
Find the general solution of $\frac{d y}{d x} = - \frac{x}{y}$ , $y \neq = 0$ , and describe the family of solution curves. Find the equation of the curve that passes through $(5, - 12)$ and state its radius.
[5 marks]
Answer
C3
Given that $y = \frac{1}{2}$ when $x = 0$ , solve $\frac{d y}{d x} = y^{2} e^{- x}$ , giving $y$ in terms of $x$ . State what happens to $y$ as $x$ becomes large.
[6 marks]
Answer
C4
Ben solves $\frac{d y}{d x} = 3 (y - 2)$ , given that $y = 1$ when $x = 0$ . He writes " $ln (y - 2) = 3 x + c$ " and then says: "At $x = 0$ this needs $ln (- 1)$ , which does not exist, so there is no solution." Explain his mistake and find the particular solution.
[5 marks]
Answer

Exam-style questions

Edexcel exam style

D1
$f (x) = \frac{4 x + 5}{( x + 2 ) ( 2 x + 1 )}$ , $x \geq 0$ .
(a)
Express $f (x)$ in partial fractions.
[3 marks]
Answer
(b)
Given that $y = 6$ when $x = 0$ , solve $\frac{d y}{d x} = y f (x)$ , $y > 0$ . Give $y$ as a polynomial in $x$ .
[5 marks]
Answer
(c)
Find the value of $x$ for which $y = 27$ .
[2 marks]
Answer
Total for D1: 10 marks

D2
Priya solves $\frac{d y}{d x} = \frac{x y}{x ^{2} + 4}$ , $y > 0$ , given that $y = 2$ when $x = 0$ . She writes: " $ln y = ln (x^{2} + 4) + c$ , so $y = x^{2} + 4 + c$ . At $x = 0$ , $2 = 4 + c$ , so $y = x^{2} + 2$ ."
Priya makes two mistakes.
(a)
Explain her first mistake.
[2 marks]
(b)
Explain her second mistake.
[2 marks]
(c)
Find the correct particular solution.
[4 marks]
Answer
(d)
Show that the correct solution curve passes through $(5, 3)$ , and that Priya's curve does not.
[2 marks]
Total for D2: 10 marks
D3
Given that $y = 1$ when $x = 0$ , solve $\frac{d y}{d x} = e^{2 x - y}$ , giving $y$ in terms of $x$ . Hence find the exact value of $x$ for which $y = 2$ .
[8 marks]
Answer
D4
A curve satisfies $\frac{d y}{d x} = \frac{4 cos ^{2} x}{y}$ , $y > 0$ , and passes through $(0, 2)$ .
(a)
Show that $4 cos^{2} x = 2 + 2 cos 2 x$ .
[2 marks]
(b)
Hence find $y^{2}$ in terms of $x$ .
[5 marks]
Answer
(c)
Find $y$ when $x = \frac{π}{4}$ , giving your answer to $3$ significant figures.
[2 marks]
Answer
Total for D4: 9 marks
D5
This question is about the differential equation $\frac{d y}{d x} = \frac{x}{y x ^{2} + 5}$ , $y > 0$ .
(a)
Use the substitution $u = x^{2} + 5$ to find $\int \frac{x}{x ^{2} + 5} d x$ .
[3 marks]
Answer
(b)
Given that $y = 3$ when $x = 2$ , find $y^{2}$ in terms of $x$ .
[4 marks]
Answer
(c)
Find the exact positive value of $x$ for which $y = 11$ .
[2 marks]
Answer
Total for D5: 9 marks
D6
This question is about the differential equation $\frac{d y}{d x} = y tan x$ , for $- \frac{π}{2} < x < \frac{π}{2}$ .
(a)
Find the general solution, in the form $y = f (x)$ .
[4 marks]
Answer
(b)
Find the particular solution through $(0, 3)$ and the exact value of $y$ on it when $x = \frac{π}{3}$ .
[2 marks]
Answer
(c)
Explain why this solution cannot be extended to $x = \frac{π}{2}$ .
[2 marks]
Total for D6: 8 marks

Extension

Stretch

No route is given. The hints are at the end of the sheet. Use one at a time.

E1
The curve $C$ satisfies $\frac{d y}{d x} = e^{x + y}$ and passes through the origin. Find $y$ in terms of $x$ . Show that $C$ has a vertical asymptote and find its equation. Find the exact $x$ -coordinate of the point on $C$ where the gradient is $4$ .
[10 marks]
Hint 1 · What to try
$e^{x + y} = e^{x} e^{y}$ , so the variables separate: $\int e^{- y} d y = \int e^{x} d x$ .
Hint 2 · The first line
You should reach $e^{- y} = 2 - e^{x}$ . What happens to $y$ as $e^{x}$ gets close to $2$ ?
Hint 3 · The full method
The gradient is $e^{x} e^{y}$ , and $e^{y} = \frac{1}{2 - e ^{x}}$ . Set the gradient equal to $4$ and solve for $e^{x}$ .

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can separate the variables and integrate both sides, putting one constant on one side.
I can take exponentials correctly, so the constant added to ln\|y\| becomes a constant A that multiplies, not one that is added.
I can use a given point to find a particular solution, and leave it in a clear form.
I can use partial fractions, integration by parts and trigonometric identities to integrate each side.

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