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
𝑑𝑥 2
+𝑏
𝑑𝑦
𝑑𝑥
+ 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.
Summary of steps to solve:
Write the Auxillary Equation (AE)
Solve AE to find the Complementary Function (CF)
Find the Particular Integral (PI)
Find the complete general solution
Find the Particular Solution (if applicable)
Particular Integral
𝒂𝒎𝟐 + 𝒃𝒎 + 𝒄 = 𝟎
e.g. 𝒚 = 𝑨𝒆𝒎𝟏 𝒙 + 𝑩𝒆𝒎𝟐 𝒙
e.g. 𝒚 = 𝑨𝒆𝒎𝟏 𝒙 + 𝑩𝒆𝒎𝟐 𝒙 + X
CF + PI
Some things you’ll need to know:
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)
Special cases of the Complementary Function:
𝑑𝑦
There is a quick solution for the CF if the equation has no 𝑑𝑥 term:
Equation:
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑥 2
Complementary Function
+ 𝑛2 𝑦 = 0
𝑦 = 𝐴𝑐𝑜𝑠(𝑛𝑥) + 𝐵𝑠𝑖𝑛(𝑛𝑥)
− 𝑛2 𝑦 = 0
𝑦 = 𝐴𝑐𝑜𝑠ℎ(𝑛𝑥) + 𝐵𝑠𝑖𝑛ℎ(𝑛𝑥)
Useful general forms for the Right Hand Side:
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 as before.
© H Jackson 2013 / Academic Skills
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Examples:
1. If the RHS of the equation is 0, then you only need to find the CF – that will be the final
answer.
e.g. Solve the differential equation
𝒅𝟐 𝒚
𝒅𝒙𝟐
−𝟓
𝒅𝒚
𝒅𝒙
+ 𝟔𝒚 = 𝟎
𝑚2 − 5𝑚 + 6 = 0
(𝑚 − 2)(𝑚 − 3) = 0
∴ 𝑚1 = 2, 𝑚2 = 3
Step 1: Use the coefficients to write the AE:
Step 2: Solve AE to find CF:
Real & different roots
so we use the 1st form
of the CF.
𝒚 = 𝑨𝒆𝟐𝒙 + 𝑩𝒆𝟑𝒙
Complementary Function:
If you were given initial conditions you could also find A and B but if not this is the final
answer to the question.
2. If the RHS is a function then you need to do all the steps and find the Complete Solution.
𝒅𝟐 𝒚
𝒅𝒚
e.g. Solve the differential equation 𝒅𝒙𝟐 − 𝟑 𝒅𝒙 + 𝟐𝒚 = 𝟐𝒙𝟐 − 𝟏𝟎𝒙 + 𝟏𝟎, given that when
𝒙 = 𝟎, 𝒚 = 𝟑 𝒂𝒏𝒅
𝒅𝒚
𝒅𝒙
=𝟕
Step 1: Use the coefficients to write the AE:
Step 2: Solve AE to find CF:
𝑚2 − 3𝑚 + 2 = 0
(𝑚 − 2)(𝑚 − 1) = 0
∴ 𝑚1 = 2, 𝑚2 = 1
Real & different roots
so we use the 1st form
of the CF.
𝒚 = 𝑨𝒆𝟐𝒙 + 𝑩𝒆𝒙
Complementary Function:
This time there is a function on the right hand side (RHS) so we continue on to find the PI.
We need to decide which general form of the RHS to use (by looking at the table above).
Step 3: Find the PI
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 to get:
𝑑𝑥
= 2𝐶𝑥 + 𝐷
𝑑2 𝑦
𝑑𝑥 2
= 2𝐶
Now substitute the above into the original equation:
Original:
𝑑2 𝑦
𝑑𝑥 2
𝑑𝑦
− 3 𝑑𝑥
+ 2𝑦
= 2𝑥 2 − 10𝑥 + 10
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
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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
Nos:
𝒚 = 𝒙𝟐 − 𝟐𝒙 + 𝟏
Therefore, Particular Integral
Step 4: Write the complete general solution:
The complete General Solution is CF+PI
𝒚 = 𝑨𝒆𝟐𝒙 + 𝑩𝒆𝒙 + 𝒙𝟐 − 𝟐𝒙 + 𝟏
So we have:
Step 5: Find the particular solution:
In this question we have been given the initial conditions (when 𝑥 = 0, 𝑦 = 3 𝑎𝑛𝑑
we can continue on to find A and B and write the particular solution.
So far we know that:
𝑑𝑥
= 7) so
𝑦 = 𝐴𝑒 2𝑥 + 𝐵𝑒 𝑥 + 𝑥 2 − 2𝑥 + 1
𝑑𝑦
Therefore:
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 / Academic Skills
𝒚 = 𝟕𝒆𝟐𝒙 − 𝟓𝒆𝒙 + 𝒙𝟐 − 𝟐𝒙 + 𝟏
3
e.g. Solve the differential equation
𝒅𝟐 𝒚
𝒅𝒙𝟐
−𝟕
Step 1: Use the coefficients to write the AE:
Step 2: Solve AE to find CF:
𝒅𝒚
𝒅𝒙
+ 𝟏𝟐𝒚 = 𝟐𝒔𝒊𝒏𝟑𝒙
𝑚2 − 7𝑚 + 12 = 0
(𝑚 − 3)(𝑚 − 4) = 0
∴ 𝑚1 = 3, 𝑚2 = 4
Real & different roots
so we use the 1st form
of the CF.
𝒚 = 𝑨𝒆𝟑𝒙 + 𝑩𝒆𝟒𝒙
Complementary Function:
Step 3: Find the PI
Looking at the RHS, 𝑓(𝑥) = 2𝑠𝑖𝑛3𝑥, so we assume the general form:
𝑦 = 𝐶𝑐𝑜𝑠3𝑥 + 𝐷𝑠𝑖𝑛3𝑥
Differentiate to get:
𝑑𝑦
𝑑𝑥
= −3𝐶𝑠𝑖𝑛3𝑥 + 3𝐷𝑐𝑜𝑠3𝑥
𝑑2 𝑦
𝑑𝑥 2
= −9𝐶𝑐𝑜𝑠3𝑥 − 9𝐷𝑠𝑖𝑛3𝑥
The trick here is to simplify and say:
𝑑2 𝑦
𝑑𝑥 2
= −9(𝐶𝑐𝑜𝑠3𝑥 + 𝐷𝑠𝑖𝑛3𝑥) = −9𝑦
𝑑2 𝑦
𝑑2 𝑦
First substitute 𝑑𝑥 2 = −9𝑦 into the original equation:
𝑑𝑥 2
𝑑𝑦
− 7 𝑑𝑥 + 12𝑦 = 2𝑠𝑖𝑛3𝑥
𝑑𝑦
−9𝑦 − 7 𝑑𝑥 + 12𝑦 = 2𝑠𝑖𝑛3𝑥
3𝑦 − 7
𝑑𝑦
𝑑𝑥
= 2𝑠𝑖𝑛3𝑥
𝑑𝑦
Next substitute 𝑦 and 𝑑𝑥 in:
→ 3(𝐶𝑐𝑜𝑠3𝑥 + 𝐷𝑠𝑖𝑛3𝑥) − 7(−3𝐶𝑠𝑖𝑛3𝑥 + 3𝐷𝑐𝑜𝑠3𝑥) = 2𝑠𝑖𝑛3𝑥
→ 3𝐶𝑐𝑜𝑠3𝑥 + 3𝐷𝑠𝑖𝑛3𝑥 + 21𝐶𝑠𝑖𝑛3𝑥 − 21𝐷𝑐𝑜𝑠3𝑥 = 2𝑠𝑖𝑛3𝑥
→ (3𝐶 − 21𝐷)𝑐𝑜𝑠3𝑥 + (21𝐶 + 3𝐷)𝑠𝑖𝑛3𝑥 = 2𝑠𝑖𝑛3𝑥
Equate coefficients of 𝑠𝑖𝑛3𝑥:
Equate coefficients of 𝑐𝑜𝑠3𝑥:
21𝐶 + 3𝐷 = 2
3𝐶 − 21𝐷 = 0
Solve these equations simultaneously to find:
Particular Integral
𝟕
7
1
𝐶 = 75 and 𝐷 = 75
𝟏
𝒚 = 𝟕𝟓 𝒄𝒐𝒔𝟑𝒙 + 𝟕𝟓 𝒔𝒊𝒏𝟑𝒙
Step 4: Write the complete general solution:
The complete General Solution is CF+PI
So we have:
𝟕
𝟏
𝒚 = 𝑨𝒆𝟑𝒙 + 𝑩𝒆𝟒𝒙 + 𝟕𝟓 𝒄𝒐𝒔𝟑𝒙 + 𝟕𝟓 𝒔𝒊𝒏𝟑𝒙
n.b. if we had been given initial conditions we could now find A and B.
© H Jackson 2013 / Academic Skills
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