First Order Differential Equations

Solving differential equations using separation of variables

SPEC 8.7 • INTEGRATION

What is a Differential Equation?

Definition

A differential equation is an equation that contains one or more derivatives of a function.

The order of a differential equation is determined by the highest derivative present:

OrderHighest DerivativeExample

First Order

\frac{d y}{d x}

\frac{d y}{d x} = x y

Second Order

\frac{d ^{2} y}{d x ^{2}}

\frac{d ^{2} y}{d x ^{2}} + y = 0

A Level Focus

At A Level, you solve first order differential equations using separation of variables.

Separation of Variables

For equations of the form

\frac{d y}{d x} = f (x) \cdot g (y)

, we separate and integrate:

Method

\frac{d y}{d x} = f (x) \cdot g (y) ⟹ \int \frac{1}{g ( y )} d y = \int f (x) d x

Critical Rule

NEVER add or subtract terms to move them to the other side!

You can only multiply or divide to separate variables. Think of

d y

and

d x

as being multiplied — rearrange by multiplication only.

The Constant

Add the constant of integration

+ C

to one side only (typically the RHS). This gives the general solution — a family of curves, one for each value of

C

.

Using ln k as the Constant

When both sides of your integral result in

ln

terms, use

ln k

instead of

C

:

Why This Works

If we have:

ln ∣ y ∣ = ln ∣ x ∣ + C

Let

C = ln k

:

ln ∣ y ∣ = ln ∣ x ∣ + ln k = ln ∣ k x ∣

Therefore:

y = k x

— much cleaner!

When to Use ln k

Both integrals give

ln

terms

You want to simplify using log rules

The final answer should be in the form

y = ...

without

ln

Log Rules Reminder

ln a + ln b = ln (ab)

and

ln a - ln b = ln (\frac{a}{b})

Worked Example01

Find the general solution of $\frac{d y}{d x} = x y$

1

Separate variables:

\frac{1}{y} d y = x d x

2

Integrate both sides:

\int \frac{1}{y} d y = \int x d x

3

ln ∣ y ∣ = \frac{x ^{2}}{2} + C

Answer: $ln ∣ y ∣ = \frac{x ^{2}}{2} + C$

Worked Example02

Find the general solution of $\frac{d y}{d x} = y tan x$

1

Separate variables:

\frac{1}{y} d y = tan x d x

2

Integrate:

\int \frac{1}{y} d y = \int tan x d x

3

ln ∣ y ∣ = - ln ∣ cos x ∣ + C = ln ∣ sec x ∣ + C

Answer: $ln ∣ y ∣ = ln ∣ sec x ∣ + C$

Worked Example03

Find the general solution of $\frac{d y}{d x} = \frac{2 x + 1}{2 y - 1}$

1

Separate variables:

2 y - 1 d y = 2 x + 1 d x

2

Rewrite:

(2 y - 1)^{\frac{1}{2}} d y = (2 x + 1)^{\frac{1}{2}} d x

3

Integrate using reverse chain rule on both sides:

4

LHS:

\int (2 y - 1)^{\frac{1}{2}} d y = \frac{( 2 y - 1 ) ^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{2} = \frac{( 2 y - 1 ) ^{\frac{3}{2}}}{3}

5

RHS:

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

6

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

Answer: $(2 y - 1)^{\frac{3}{2}} = (2 x + 1)^{\frac{3}{2}} + C$

Finding Particular Solutions

A particular solution is found by using an initial condition to determine the value of the constant

C

.

Method

Find the general solution (with constant

C

)

Substitute the given condition (e.g.,

y = y_{0}

when

x = x_{0}

)

Solve for

C

Write the particular solution

Notation

"

y = 3

when

x = 0

" can also be written as "

y (0) = 3

" — both mean the same thing.

Worked Example04

Solve $\frac{d y}{d x} = 2 x y$ given that $y = 3$ when $x = 0$ .

1

Separate:

\frac{1}{y} d y = 2 x d x

2

Integrate:

ln ∣ y ∣ = x^{2} + C

3

General solution:

y = A e^{x^{2}}

4

Apply condition: When

x = 0

,

y = 3

:

3 = A e^{0} = A

5

So

A = 3

Answer: $y = 3 e^{x^{2}}$

Worked Example05

Solve $\frac{d y}{d x} = \frac{y ^{2}}{x}$ given that $y = 1$ when $x = 1$ .

1

Separate:

\frac{1}{y ^{2}} d y = \frac{1}{x} d x

, i.e.,

y^{- 2} d y = \frac{1}{x} d x

2

Integrate:

\int y^{- 2} d y = \int \frac{1}{x} d x

3

- y^{- 1} = ln ∣ x ∣ + C

, i.e.,

- \frac{1}{y} = ln ∣ x ∣ + C

4

Condition:

x = 1, y = 1

:

- 1 = ln 1 + C = 0 + C

, so

C = - 1

5

- \frac{1}{y} = ln x - 1

, so

\frac{1}{y} = 1 - ln x

Answer: $y = \frac{1}{1 - l n x}$

More Complex Separable DEs

Some differential equations require additional techniques:

Common Situations

Factorising first:

\frac{d y}{d x} = y - y^{2} = y (1 - y)

Rearranging:

x \frac{d y}{d x} = y^{2}

becomes

\frac{d y}{d x} = \frac{y ^{2}}{x}

Partial fractions: When

\frac{1}{g ( y )}

needs decomposing

Trig integrals:

sec^{2} y

,

csc y

, etc.

Implicit Solutions

Sometimes you cannot rearrange to get

y = ...

explicitly. In such cases, leave your answer in implicit form.

Worked Example06

Solve $x \frac{d y}{d x} = y (1 - y)$ given that $y = \frac{1}{2}$ when $x = 1$ . Express $y$ in terms of $x$ .

1

Separate:

\frac{1}{y ( 1 - y )} d y = \frac{1}{x} d x

2

Partial fractions:

\frac{1}{y ( 1 - y )} = \frac{1}{y} + \frac{1}{1 - y}

3

Integrate:

ln ∣ y ∣ - ln ∣1 - y ∣ = ln ∣ x ∣ + C

4

Use

C = ln k

:

ln \frac{y}{1 - y} = ln ∣ k x ∣

, so

\frac{y}{1 - y} = k x

5

Condition:

x = 1, y = \frac{1}{2}

:

\frac{1/2}{1/2} = k

, so

k = 1

6

\frac{y}{1 - y} = x ⟹ y = x (1 - y) ⟹ y (1 + x) = x

Answer: $y = \frac{x}{1 + x}$

Worked Example07

Solve $sec^{2} y \frac{d y}{d x} = e^{x}$ given that $y = 0$ when $x = 0$ .

1

Separate:

sec^{2} y d y = e^{x} d x

2

Integrate:

\int sec^{2} y d y = \int e^{x} d x

3

tan y = e^{x} + C

4

Condition:

x = 0, y = 0

:

tan 0 = e^{0} + C

, so

0 = 1 + C

,

C = - 1

5

tan y = e^{x} - 1

Answer: $y = arctan (e^{x} - 1)$

Worked Example08

Solve $\frac{d y}{d x} = e^{x - y}$ given that $y = 0$ when $x = 0$ .

1

Rewrite:

\frac{d y}{d x} = e^{x} \cdot e^{- y} = \frac{e ^{x}}{e ^{y}}

2

Separate:

e^{y} d y = e^{x} d x

3

Integrate:

\int e^{y} d y = \int e^{x} d x

4

e^{y} = e^{x} + C

5

Condition:

x = 0, y = 0

:

e^{0} = e^{0} + C

, so

1 = 1 + C

,

C = 0

6

e^{y} = e^{x}

, so

y = x

Answer: $y = x$

Worked Example09

Solve $\frac{d y}{d x} = \frac{x e ^{x}}{e ^{y}}$ given that $y = 0$ when $x = 1$ .

1

Separate:

e^{y} d y = x e^{x} d x

2

LHS:

\int e^{y} d y = e^{y}

3

RHS:

\int x e^{x} d x

— use integration by parts with

u = x

,

\frac{d v}{d x} = e^{x}

4

\int x e^{x} d x = x e^{x} - \int e^{x} d x = x e^{x} - e^{x} = e^{x} (x - 1)

5

So:

e^{y} = e^{x} (x - 1) + C

6

Condition:

x = 1, y = 0

:

e^{0} = e^{1} (1 - 1) + C

, so

1 = 0 + C

,

C = 1

7

e^{y} = e^{x} (x - 1) + 1

Answer: $y = ln (e^{x} (x - 1) + 1)$

10

Worked Example

Solve $(1 + x^{2}) \frac{d y}{d x} = x y$ given that $y = 2$ when $x = 0$ .

1

Separate:

\frac{1}{y} d y = \frac{x}{1 + x ^{2}} d x

2

Integrate:

\int \frac{1}{y} d y = \int \frac{x}{1 + x ^{2}} d x

3

RHS: Let

u = 1 + x^{2}

,

d u = 2 x d x

, so

\int \frac{x}{1 + x ^{2}} d x = \frac{1}{2} ln (1 + x^{2})

4

ln ∣ y ∣ = \frac{1}{2} ln (1 + x^{2}) + C = ln 1 + x^{2} + ln k

(using

C = ln k

)

5

y = k 1 + x^{2}

. Condition:

x = 0, y = 2

:

2 = k 1 = k

Answer: $y = 2 1 + x^{2}$

Practice

Question 1: Find the general solution of

\frac{d y}{d x} = y cos x

1

Separate:

\frac{1}{y} d y = cos x d x

2

Integrate:

ln ∣ y ∣ = sin x + C

Answer: $ln ∣ y ∣ = sin x + C$

Practice

Question 2: Solve

\frac{d y}{d x} = \frac{x}{y}

given that

y = 2

when

x = 0

.

1

Separate:

y d y = x d x

2

Integrate:

\frac{y ^{2}}{2} = \frac{x ^{2}}{2} + C

3

General:

y^{2} = x^{2} + 2 C

. Let

k = 2 C

:

y^{2} = x^{2} + k

4

Condition:

x = 0, y = 2

:

4 = 0 + k

, so

k = 4

Answer: $y = x^{2} + 4$ (taking positive root)

Practice

Question 3: Solve

\frac{d y}{d x} = \frac{3 y}{x}

given that

y = 8

when

x = 2

.

1

Separate:

\frac{1}{y} d y = \frac{3}{x} d x

2

Integrate:

ln ∣ y ∣ = 3 ln ∣ x ∣ + C

3

Use

C = ln k

:

ln ∣ y ∣ = ln ∣ x^{3} ∣ + ln k = ln ∣ k x^{3} ∣

4

So

y = k x^{3}

. Condition:

x = 2, y = 8

:

8 = k (8)

, so

k = 1

Answer: $y = x^{3}$

Practice

Question 4: Solve

\frac{d y}{d x} = \frac{x c o s x}{y}

given that

y = 1

when

x = 0

.

1

Separate:

y d y = x cos x d x

2

LHS:

\int y d y = \frac{y ^{2}}{2}

3

RHS:

\int x cos x d x

— by parts:

u = x

,

\frac{d v}{d x} = cos x

4

\int x cos x d x = x sin x - \int sin x d x = x sin x + cos x

5

So:

\frac{y ^{2}}{2} = x sin x + cos x + C

6

Condition:

x = 0, y = 1

:

\frac{1}{2} = 0 + 1 + C

, so

C = - \frac{1}{2}

Answer: $\frac{y ^{2}}{2} = x sin x + cos x - \frac{1}{2}$ or $y^{2} = 2 x sin x + 2 cos x - 1$

Practice

Question 5: Solve

cos^{2} y \frac{d y}{d x} = sin^{2} x

given that

y = 0

when

x = 0

.

1

Separate:

cos^{2} y d y = sin^{2} x d x

2

Use identity:

cos^{2} y = \frac{1 + c o s 2 y}{2}

and

sin^{2} x = \frac{1 - c o s 2 x}{2}

3

\int \frac{1 + c o s 2 y}{2} d y = \int \frac{1 - c o s 2 x}{2} d x

4

\frac{1}{2} (y + \frac{s i n 2 y}{2}) = \frac{1}{2} (x - \frac{s i n 2 x}{2}) + C

5

Simplify:

y + \frac{s i n 2 y}{2} = x - \frac{s i n 2 x}{2} + C

6

Condition:

x = 0, y = 0

:

0 + 0 = 0 - 0 + C

, so

C = 0

Answer: $2 y + sin 2 y = 2 x - sin 2 x$

Summary

Key Steps for Separation of Variables

1

Separate — get all

y

terms with

d y

, all

x

terms with

d x

(multiply/divide only!)

2

Integrate both sides

3

Add

+ C

to one side for general solution (use

ln k

if both sides are

ln

)

4

Apply condition if given to find particular solution

Common Integrals You'll Need

\int \frac{1}{y} d y = ln ∣ y ∣

\int y^{n} d y = \frac{y ^{n + 1}}{n + 1}

\int e^{y} d y = e^{y}

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

Exam Tip

Always check your answer by differentiating — substituting back should give the original DE!