2nd Order Differentials Summary
Academic Skills Advice
A 2nd order differential equation is one that has a 2nd derivative in it. For example:
𝑎
𝑑2𝑦
𝑑𝑦
𝑑𝑥
𝑑𝑥
+𝑏
2
+ 𝑐𝑦 = 𝑓(𝑥)
To solve this equation you would need to find 𝑦, which is a function of 𝑥. This summary
demonstrates a method to solve the equation. Examples are worked through overleaf.
Summary of steps to solve:
Step 1:
Step 2:
Step 3:
Step 4:
Solve the left hand side
Solve the right hand side
Add them together
Use initial conditions
Some jargon:
You will create an Auxiliary Equation (AE) 𝒂𝒎𝟐 + 𝒃𝒎 + 𝒄 = 𝟎
to help find the Complementary Function (CF)
Right hand side: You will find the Particular Integral (PI)
Add together:
You have found the complete general solution (CF+PI)
Initial conditions: You will find the Particular solution.
Left hand side:
When solving the left hand side, how you write the Complementary Function will
depend on the roots of the Auxiliary Equation that you find:
Type of roots:
Real & different
Real & equal
Complex (𝑚 = 𝛼 ± 𝛽𝑖)
Complementary Function:
𝑦 = 𝐴𝑒 𝑚1 𝑥 + 𝐵𝑒 𝑚2 𝑥
𝑦 = 𝑒 𝑚1 𝑥 (𝐴 + 𝐵𝑥)
𝑦 = 𝑒 𝛼𝑥 (𝐴𝑐𝑜𝑠𝛽𝑥 + 𝐵𝑠𝑖𝑛𝛽𝑥)
A short cut:
If your equation looks like this:
𝑑2 𝑦
𝑎 𝑑𝑥 2 + 𝑐𝑦 = 0
𝑐
𝑐
You can just write the complementary function straight down: 𝑦 = 𝐴𝑐𝑜𝑠 (√𝑎 𝑥) + 𝐵𝑠𝑖𝑛(√𝑎 𝑥)
(Don’t worry if you can’t remember this – just stick to the 3 above)
To solve the Right Hand Side we need to identify the general form:
If:
𝑓(𝑥) = 𝑘
𝑓(𝑥) = 𝑘𝑥
𝑓(𝑥) = 𝑘𝑥 2
𝑓(𝑥) = 𝑘𝑠𝑖𝑛𝑥 𝑜𝑟 𝑘𝑐𝑜𝑠𝑥
𝑓(𝑥) = 𝑘𝑠𝑖𝑛ℎ𝑥 𝑜𝑟 𝑘𝑐𝑜𝑠ℎ𝑥
𝑓(𝑥) = 𝑒 𝑘𝑥
Assume:
𝑦=𝐶
𝑦 = 𝐶𝑥 + 𝐷
𝑦 = 𝐶𝑥 2 + 𝐷𝑥 + 𝐸
𝑦 = 𝐶𝑐𝑜𝑠𝑥 + 𝐷𝑠𝑖𝑛𝑥
𝑦 = 𝐶𝑐𝑜𝑠ℎ𝑥 + 𝐷𝑠𝑖𝑛ℎ𝑥
𝑦 = 𝐶𝑒 𝑘𝑥
n.b. If the general form of the RHS is already included in the CF then multiply the assumed
general form by 𝑥, then continue with this new general form.
H Jackson 2013 / 2016 / Academic Skills
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Step 1: Solve the left hand side



Use the coefficients to write the Auxiliary Equation (AE) 𝒂𝒎𝟐 + 𝒃𝒎 + 𝒄 = 𝟎
Solve AE to find 𝑚1 and 𝑚2
Choose the correct Complementary Function (CF)
E.g. 1 Solve the differential equation
𝒅𝟐 𝒚
𝒅𝒙𝟐
−𝟓
𝒅𝒚
𝒅𝒙
+ 𝟔𝒚 = 𝟎
Use the coefficients to write the AE:
Solve AE to find 𝑚1 and 𝑚2 :
𝒎𝟐 − 𝟓𝒎 + 𝟔 = 𝟎
(𝑚 − 2)(𝑚 − 3) = 0
∴ 𝑚1 = 2, 𝑚2 = 3
CF for “real & different roots”:
𝒚 = 𝑨𝒆𝟐𝒙 + 𝑩𝒆𝟑𝒙
The right hand side = 0 and there are no given initial conditions so this is the final answer to
the question.
E.g. 2 Solve the differential equation
𝒙 = 𝟎, 𝒚 = −𝟏 𝒂𝒏𝒅
𝒅𝒚
𝒅𝒙
𝒅𝟐 𝒚
𝒅𝒙𝟐
−𝟒
𝒅𝒚
𝒅𝒙
+ 𝟑𝒚 = 𝟎, given that when
=𝟏
Use the coefficients to write the AE:
Solve AE to find 𝑚1 and 𝑚2 :
𝒎𝟐 − 𝟒𝒎 + 𝟑 = 𝟎
(𝑚 − 3)(𝑚 − 1) = 0
∴ 𝑚1 = 3, 𝑚2 = 1
CF for “real & different roots”:
𝒚 = 𝑨𝒆𝟑𝒙 + 𝑩𝒆𝒙
The right hand side = 0 but this time there are given initial conditions so go straight to step 4
and use the initial conditions to find A and B.
Step 4: Initial Conditions



𝑑𝑦
Differentiate to find 𝑑𝑥
Substitute initial conditions in
Solve to find A and B
E.g.2 cont... We have found the CF:
Differentiate:
Substitute 𝑥 = 0, 𝑦 = −1
𝑑𝑦
Substitute 𝑥 = 0, 𝑑𝑥 = 1
𝒚 = 𝑨𝒆𝟑𝒙 + 𝑩𝒆𝒙
𝑑𝑦
𝑑𝑥
= 3𝐴𝑒 3𝑥 + 𝐵𝑒 𝑥
−1 = 𝐴𝑒 0 + 𝐵𝑒 0
1 = 3𝐴𝑒 0 + 𝐵𝑒 0
−1 = 𝐴 + 𝐵
1 = 3𝐴 + 𝐵
𝐴 = 1, 𝐵 = −2
Solve simultaneously:
Final answer:
→
→
𝒚 = 𝒆𝟑𝒙 − 𝟐𝒆𝒙
It’s a good idea to practice a few of the above type of questions before moving on to
equations where 𝑅𝐻𝑆 ≠ 0 (i.e. when there is something on the right hand side).
H Jackson 2013 / 2016 / Academic Skills
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E.g. 3 Solve the differential equation
given that when 𝒙 = 𝟎, 𝒚 = 𝟑 𝒂𝒏𝒅
𝒅𝒚
𝒅𝒙
Step 1: Solve the LHS:
Use the coefficients to write the AE:
Solve AE to find 𝑚1 and 𝑚2 :
𝒅𝟐 𝒚
𝒅𝒙𝟐
−𝟑
𝒅𝒚
𝒅𝒙
+ 𝟐𝒚 = 𝟐𝒙𝟐 − 𝟏𝟎𝒙 + 𝟏𝟎,
=𝟕
𝑚2 − 3𝑚 + 2 = 0
(𝑚 − 2)(𝑚 − 1) = 0
∴ 𝑚1 = 2, 𝑚2 = 1
𝒚 = 𝑨𝒆𝟐𝒙 + 𝑩𝒆𝒙
CF for “real & different roots”:
Step 2: Solve the right hand side



Write down the correct assumption
Differentiate twice then substitute into original
Tidy up and find C, D etc.
Looking at the RHS, 𝑓(𝑥) = 2𝑥 2 − 10𝑥 + 10, so (from the table) we assume the general form:
𝒚 = 𝑪𝒙𝟐 + 𝑫𝒙 + 𝑬
𝑑𝑦
Differentiate once:
𝑑𝑥
= 2𝐶𝑥 + 𝐷
𝑑2 𝑦
Differentiate again:
𝑑𝑥 2
= 2𝐶
Substitute into original equation:
Original:
𝑑2 𝑦
𝑑𝑥 2
𝑑𝑦
− 3 𝑑𝑥
= 2𝑥 2 − 10𝑥 + 10
+ 2𝑦
Becomes:
2𝐶 − 3(2𝐶𝑥 + 𝐷) + 2(𝐶𝑥 2 + 𝐷𝑥 + 𝐸) = 2𝑥 2 − 10𝑥 + 10
Tidy up to get:
2𝐶 − 6𝐶𝑥 − 3𝐷 + 2𝐶𝑥 2 + 2𝐷𝑥 + 2𝐸 = 2𝑥 2 − 10𝑥 + 10
Now we need to solve to find C, D & E. The easiest way to do this is to equate coefficients:
Equate Coefficients to find 𝑪, 𝑫 & 𝑬:
𝑥2:
2𝐶 = 2
∴𝐶=1
𝑥:
−6𝐶 + 2𝐷 = −10
−6 + 2𝐷 = −10
2𝐷 = −4
∴ 𝐷 = −2
2𝐶 − 3𝐷 + 2𝐸 = 10
2 + 6 + 2𝐸 = 10
2𝐸 = 2
∴𝐸=1
Constants:
Therefore, Particular Integral
H Jackson 2013 / 2016 / Academic Skills
𝒚 = 𝒙𝟐 − 𝟐𝒙 + 𝟏
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Step 3: Add them together

𝑦 = 𝐶𝐹 + 𝑃𝐼
(complementary function + particular integral)
Complete General Solution:
𝒚 = 𝑨𝒆𝟐𝒙 + 𝑩𝒆𝒙 + 𝒙𝟐 − 𝟐𝒙 + 𝟏
Step 4: Initial Conditions
Substitute in the initial conditions to get the final answer:
We’re given that when 𝑥 = 0, 𝑦 = 3 𝑎𝑛𝑑
So far we know that:
𝑑𝑥
=7
𝑦 = 𝐴𝑒 2𝑥 + 𝐵𝑒 𝑥 + 𝑥 2 − 2𝑥 + 1
𝑑𝑦
Differentiate:
𝑑𝑦
𝑑𝑥
= 2𝐴𝑒 2𝑥 + 𝐵𝑒 𝑥 + 2𝑥 − 2
Substitute initial conditions into the above:
When 𝑥 = 0, 𝑦 = 3:
𝑑𝑦
When 𝑥 = 0, 𝑑𝑥 = 7:
3=𝐴+𝐵+1
→
𝐴+𝐵 =2
7 = 2𝐴 + 𝐵 − 2
→
2𝐴 + 𝐵 = 9
Solve simultaneously to find that, 𝐴 = 7 and 𝐵 = −5
We now have the particular solution:
H Jackson 2013 / 2016 / Academic Skills
𝒚 = 𝟕𝒆𝟐𝒙 − 𝟓𝒆𝒙 + 𝒙𝟐 − 𝟐𝒙 + 𝟏
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