FP2
Last modified: 19 th January 2016

Differential equations are equations which relate 𝑥 and 𝑦 with derivatives. e.g.
The rate of temperature loss is proportional to the current temperature.
𝑑𝑇 𝑑𝑡
= −𝑘𝑇
The rate of population change is
𝑃 proportional to 𝑃 1 − where 𝑃 is
𝑀 the current population and 𝑀 is the limiting size of the population
(the Verhulst-Pearl Model) 𝑑𝑃 𝑑𝑡
𝑃
= 𝑘𝑃 1 −
𝑀
Suppose 𝑥 is GDP (Gross Domestic Product).
Rate of change of GDP is proportional to current GDP.
𝑑𝑥
= 𝑘𝑥 𝑑𝑡
As you might imagine, they’re used a lot in physics and engineering, including modelling radioactive decay, mixing fluids, cooling materials and bodies falling under gravity against resistance.
A ‘first order’ differential equation means the equation contains the first derivative ( 𝑑𝑦 𝑑𝑥 second derivative or beyond.
) but not the

𝑥 and 𝑦 are said to be ‘separated’ because they are not mixed in the same expression.
𝑑𝑦 𝑑𝑥
1 𝑑𝑦
= 𝑓 𝑥 𝑔 𝑦
1 𝑑𝑥 𝑔 𝑦
1 𝑑𝑦 = 𝑓 𝑥 𝑑𝑥 𝑔 𝑦 𝑑𝑦 = 𝑓 𝑥 𝑑𝑥
Divide through by 𝑔(𝑦) and times through by 𝑑𝑥 .
Slap and integral symbol on the front!

Find general solutions to 𝑑𝑦 𝑑𝑥
= 2 𝑦 = 2𝑥 + 𝐶
Find general solutions to 𝑑𝑦 𝑑𝑥 𝑥
= − 𝑦
𝑦 𝑑𝑦 = −𝑥 𝑑𝑥
1
2 𝑥 2 𝑦 2 = −
+ 𝑦 2
?
1 𝑥 2
2
= 2𝐶
+ 𝐶
If we let 𝑟 2 𝑥 2
= 2𝐶 , we get
+ 𝑦 2 = 𝑟 2
Why is it called the general solution?
We have a ‘family’ of solutions as the
?
𝑦
So the ‘family of circles’ satisfies this differential equation.
𝑥

Find general solutions to 𝑑𝑦 𝑑𝑥
= 𝑥𝑦 + 𝑥 𝑑𝑦
= 𝑥 𝑦 + 1 𝑑𝑥
1 𝑑𝑦 = 𝑥 𝑑𝑥 𝑦 + 1
?
ln 𝑦 + 1 =
1
2 𝑥
2
+ 𝐶
1 1 𝑦 + 1 = 𝑒 2𝑥
2
+𝐶
= 𝑒 2𝑥
2 𝑒
= 𝐴𝑒 where 𝐴 = 𝑒 𝐶
1
2𝑥
2
𝐶
Find general solutions to 𝑑𝑦 𝑑𝑥
= − 𝑦 𝑥
1 𝑦 𝑑𝑦 = −
1 𝑑𝑥 𝑥 ln |𝑦| = − ln 𝑥 + 𝐶 ln 𝑥 + ln |𝑦| = 𝐶
?
|𝑥𝑦| = 𝑒 𝐶 = 𝐴
𝐴 𝑦 = ± 𝑥
(a family of rectangular hyperbolae)

Just like at the end of C1 integration, we can fix to one particular solution if we give some conditions.
Find the particular solutions for 𝑑𝑥 𝑑𝑡 𝑥 = 9 when 𝑡 = 0 .
= 𝑥 which satisfy the initial conditions
𝑥
−
1
2 𝑑𝑥 = 1 𝑑𝑦
2𝑥
1
2 = 𝑡 + 𝑐
2 𝑥 = 𝑡 + 𝑐
2 3 = 0 + 𝑐 → 𝑐 = 6
2 𝑥 = 𝑡 + 6
2 𝑡 + 6 𝑥 =
2
= 𝑡 + 6
4
2
?
Some people (including myself) like to sub in as soon as possible before rearranging.

We will see in a bit how to solve equations of the form 𝑑𝑦
+ 𝑃𝑦 = 𝑄 (where 𝑃 and 𝑄 𝑑𝑥 are functions of 𝑥 ). We’ll practice a particular part of this method before going for the full whack.
Find general solutions of the equation 𝑥
3 𝑑𝑦 𝑑𝑥
+ 3𝑥
2 𝑦 = sin 𝑥
We can’t separate the variables. But do you notice anything about the LHS?
𝒅
It’s 𝒅𝒙
(𝒙 ?
𝟑 𝒚) !
Quickfire Questions: 𝑑 𝑑𝑥 𝑑 𝑑𝑥 𝑥 2 𝑦 = 𝒙 𝟐 𝒅𝒚 𝒅𝒙 𝑦 sin 𝑥 = 𝐬𝐢𝐧 𝒙 𝒅𝒚 𝒅𝒙
?
+ 𝒚 𝐜𝐨𝐬 𝒙
So it appears whatever term ends up on front of the 𝑑𝑦 𝑑𝑥 in the integral.
will be on the front of the 𝑦 𝑥 𝑒
4 𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 ln 𝑥
+ 4𝑥 3 𝑦 → 𝒅 𝒅𝒙 𝒅
+ 𝑒 𝑑𝑦 𝑑𝑥 𝑥
+ 𝑦 → 𝑦 𝑥
→ 𝒅𝒙 𝒅 𝒅𝒙 𝒙
?
𝟒 𝒆 𝒙 ?
𝒚 𝒚

Find general solutions of the equation 𝑥
3 𝑑𝑦 𝑑𝑥
+ 3𝑥
2 𝑦 = sin 𝑥 𝑑 𝑑𝑥 𝑥 3 𝑥 3 𝑦 = sin 𝑥 𝑦 =
𝐶 − cos 𝑥 𝑥 3
Find general solutions of the equation
4𝑥𝑦 𝑑𝑦
+ 2𝑦 2 = 𝑥 2 𝑑𝑥
Find general solutions of the equation
1 𝑑𝑦
−
1 𝑦 = 𝑒 𝑥 𝑥 𝑑𝑥 𝑥 2 𝑑 𝑦
?
= 𝑒 𝑥 𝑑𝑥 𝑦 𝑥
= 𝑒 𝑥 𝑥 𝑦 = 𝑥(𝑒
+ 𝐶 𝑥 + 𝐶) 𝑑 𝑑𝑥
2𝑥𝑦
2
2𝑥𝑦 2
=
1
3 𝑥
?
= 𝑥
3
2
+ 𝐶
…

But what if we can’t use the product rule backwards?
Find the general solution of 𝑑𝑦 𝑑𝑥
− 4𝑦 = 𝑒 𝑥
We can multiply through by the integrating factor 𝒆 𝑷 𝒅𝒙
. This then produces an equation where we can use the previous reverse-product-rule trick (we’ll prove this in a sec).
𝐼. 𝐹. = 𝑒 −4 𝑑𝑥 −4𝑥
Then multiplying through by the integrating factor: 𝑒 −4𝑥 𝑑𝑦
− 4𝑒 −4𝑥 ?
𝑦 = 𝑒 −3𝑥 𝑑𝑥
Then we can solve in the usual way: 𝑑 𝑦𝑒 𝑑𝑥
−4𝑥 𝑦𝑒
−4𝑥 𝑦 = −
1
3 𝑒 𝑥
= 𝑒
−3𝑥
= −
1
3 𝑒
?
−3𝑥
+ 𝐶𝑒
4𝑥
+ 𝑐

Solve the general equation 𝑑𝑦 𝑑𝑥 functions of 𝑥 .
+ 𝑃𝑦 = 𝑄 , where 𝑃, 𝑄 are
Suppose 𝑓(𝑥) is the Integrating Factor. As usual we’d multiply by it: 𝑑𝑦 𝑓 𝑥 + 𝑓 𝑥 𝑃𝑦 = 𝑓 𝑥 𝑄 𝑑𝑥
If we can use the reverse product rule trick on the LHS, then it would be of the form: 𝑓 𝑥 𝑓 ′ 𝑑𝑦 ?
𝑑𝑥
Thus comparing the coefficients of the two LHSs:
?
′ 𝑥 𝑦 𝑥 = 𝑓 𝑥 𝑃
Dividing by 𝑓(𝑥) and integrating: 𝑓 ′ 𝑥 𝑓 𝑥 𝑑𝑥 = 𝑃 𝑑𝑥 ln 𝑓 𝑥
?
𝑓 𝑥 = 𝑒
𝑃 𝑑𝑥

Find the general solution of cos 𝑥 𝑑𝑦 𝑑𝑥
+ 2𝑦 sin 𝑥 = cos 4 𝑥
What shall we do first so that we have an equation like before?
𝑑𝑦
+ 2𝑦 tan 𝑥 = cos
3 𝑥 𝑑𝑥
?
𝐼𝐹 = 𝑒
2tan 𝑥 𝑑𝑥
= 𝑒 = 𝑒 ln sec
2 sec 𝑑
2 𝑥 𝑑𝑦 𝑑𝑥 𝑦 sec 𝑑𝑥 𝑦 sec 2
2
+ 2𝑦 sec 2 𝑥 = sin 𝑥 + 𝑐 𝑥 tan 𝑥 = cos 𝑥 𝑥 = cos 𝑥
?
𝑦 = cos 𝑥 sin 𝑥 + 𝑐
?
𝑥
= sec
2 𝑥
You could skip to here provided you don’t forget to multiply the
RHS by the IF.
STEP 1: Divide by anything on front of 𝑑𝑦/𝑑𝑥
STEP 2: Determine IF
STEP 3: Multiply through by IF and use product rule backwards.
STEP 4: Integrate and simplify.

FP2 June 2011 Q3
Find the general solution of the differential equation 𝑑𝑦 𝑥 + 5𝑦 = ln 𝑥
, 𝑥 > 0 𝑑𝑥 𝑥
?

Sometimes making a substitution will turn a more complex differential equation into one like we’ve seen before.
Use the substitution 𝑧 = 𝑦 𝑥 to transform the differential equation 𝑑𝑦 𝑑𝑥
= 𝑥
2
+3𝑦
2
2𝑥𝑦
, 𝑥 > 0 into a differential equation in 𝑧 and 𝑥 . By first solving this new equation, find the general solution of the original equation, giving 𝑦
2 in terms of 𝑥 .
Make 𝑦 the subject so that we can find 𝑑𝑦 𝑑𝑥 and sub 𝑦 out.
𝑧 = 𝑑𝑦 𝑑𝑥 𝑑𝑧 𝑥 𝑦 𝑥
= 𝑥 𝑑𝑥 𝑑𝑧 𝑥 = 𝑑𝑥
2𝑧
1 + 𝑧 2
→ 𝑦 = 𝑥𝑧 𝑑𝑧 𝑑𝑥
+ 𝑧 =
1
2𝑧
+ 𝑧
+ 𝑥 2
1
2
2𝑥 2
?
+ 3𝑥 𝑧 = 𝑑𝑧 =
1 𝑥
?
?
2 𝑧
2𝑧 𝑑𝑥
2 𝑧
1 + 𝑧
=
2
…
1 + 𝑧 𝑦
2
2
= 𝑥
2
= 𝐴𝑥
𝐴𝑥 − 1 ?
1
2𝑧
+
3
2 𝑧 Sub in 𝑦 and 𝑑𝑦 𝑑𝑥
Simplify to either get 𝑑𝑧 𝑑𝑥 𝑑𝑧
= 𝑓 𝑥 𝑔 𝑧 or 𝑑𝑥
+ 𝑃𝑧 = 𝑄
Sub back 𝑦 in at the end.

Use the substitution 𝑧 = 𝑦 −1 to transform the differential equation 𝑑𝑦
+ 𝑥𝑦 = 𝑥𝑦 2
, into 𝑑𝑥 a differential equation in 𝑧 and 𝑥 . Find the general solution using an integrating factor.
𝑦 = 𝑑𝑦
1
− 𝑑𝑧 𝑧 2 𝑑𝑥
+
− 𝑥𝑧 = −𝑥 𝑑𝑥
1
, 𝑧 𝑑𝑧 𝑥 𝑑𝑥
= 𝑧
𝐼𝐹 = 𝑒
−𝑥 𝑑𝑥 𝑑 𝑧𝑒 𝑧𝑒 𝑑𝑥
−
1
2𝑥
2
−
1
2𝑥
2
= 𝑒
−
1
2𝑥
2
= − 𝑥 𝑧 2
+ 𝑐 𝑧
1
2
−
1
= 𝑒
?
2𝑥
= −𝑥𝑒
2
−
1
2𝑥 𝑑𝑧 𝑑𝑥
2
1
2𝑥
2 𝑧 = 1 + 𝑐𝑒
1 𝑦 =
1 + 𝑐𝑒
1
2 𝑥 2
The 𝑥 and 𝑧 are separable here
(but factorising out 𝑥 ), so we could have done without an IF.

FP2 June 2012 Q7
(a) Show that the substitution 𝑦 = 𝑣𝑥 transforms the differential equation
3𝑥𝑦 2 𝑑𝑦
= 𝑥 3 𝐼 𝑑𝑥
+ 𝑦 3 into the differential equation
3𝑣 2 𝑥 𝑑𝑣
= 1 − 2𝑣 3 𝐼𝐼 𝑑𝑥
(b) By solving differential equation (II), find a general solution of differential equation (I) in the form 𝑦 = 𝑓 𝑥 .
(c) Given that 𝑦 = 2 at 𝑥 = 1 , find the value of 𝑑𝑦 𝑑𝑥 at 𝑥 = 1 .
a ?
b ?
c ?