1. Differential Equations
Substitution for First and Second ORDER

2. First Order vs Second Order
A first order differential equation contains a 𝑑𝑦 𝑑𝑥 term.
This is what we’ve seen so far.
A second order differential equation contains a 𝑑 2 𝑦 𝑑 𝑥 2 term.

3. Substitution
So far, we’ve seen “Separation of Variables”, and “Integrating Factors”. The next technique we will use is called “Substitution”.
When we have something that has a common combination of 𝑥 and 𝑦, we make the substitution 𝑧 = 𝑓 𝑥, 𝑦 . Often, this is a ratio or difference of 𝑥 and 𝑦.

4. Example
We can’t see anything consistent between 𝑥 and 𝑦 yet, so we divide by 𝑥 2 to find something.
𝑑𝑦 𝑑𝑥 ⋅𝑥𝑦+4 𝑥 2 + 𝑦 2 =0
𝑑𝑦 𝑑𝑥 ⋅ 𝑥𝑦 𝑥 𝑥 2 𝑥 𝑦 2 𝑥 2 =0
𝑑𝑦 𝑑𝑥 ⋅ 𝑦 𝑥 𝑦 𝑥 2 =0
𝑧= 𝑦 𝑥
𝑑𝑦 𝑑𝑥 =𝑧+𝑥 𝑑𝑧 𝑑𝑥
𝑦=𝑥𝑧
𝑧+𝑥 𝑑𝑧 𝑑𝑥 ⋅𝑧+4+ 𝑧 2 =0
𝑥𝑧 𝑑𝑧 𝑑𝑥 +4+2 𝑧 2 =0

5. I didn’t give you initial conditions, so we are, in fact, done! Yay!
Example
I didn’t give you initial conditions, so we are, in fact, done! Yay!
𝑥𝑧 𝑑𝑧 𝑑𝑥 +4+2 𝑧 2 =0
2 𝑧 2 +4= 𝑘 𝑥 4
𝑥 𝑑𝑧 𝑑𝑥 = −4 −2 𝑧 2 𝑧
2 𝑦 𝑥 = 𝑘 𝑥 4
−𝑧 2 𝑧 𝑑𝑧 = 1 𝑥 .𝑑𝑥
2 𝑦 2 𝑥 2 = 𝑘 𝑥 4 −4
𝑦 2 = 𝑘 𝑥 2 −2 𝑥 2
− 1 4 ln 2 𝑧 = ln 𝑥 +𝐶

6. Substitution for Second Order
The second order differential equations we will consider are of a very special type: they contain only 𝑥 terms, or they contain only 𝑦 terms.
We use a very specific substitution, 𝑧= 𝑑𝑦 𝑑𝑥 , and the following identity:
𝑑 2 𝑦 𝑑 𝑥 2 = 𝑑 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 𝑑𝑧 𝑑𝑥

7. Example 1 𝑥 𝑑 2 𝑦 𝑑 𝑥 2 +4 𝑑𝑦 𝑑𝑥 = 𝑥 2 𝑧= 𝑑𝑦 𝑑𝑥 𝑑𝑧 𝑑𝑥 = 𝑑 2 𝑦 𝑑 𝑥 2
This equation cannot be separated, so we have to use an integrating factor.
𝑥 𝑑 2 𝑦 𝑑 𝑥 𝑑𝑦 𝑑𝑥 = 𝑥 2
𝑧= 𝑑𝑦 𝑑𝑥
𝑑𝑧 𝑑𝑥 = 𝑑 2 𝑦 𝑑 𝑥 2
𝑥 𝑑𝑧 𝑑𝑥 +4𝑧= 𝑥 2
𝑑𝑧 𝑑𝑥 + 4𝑧 𝑥 =𝑥
𝑝(𝑥)= 4 𝑥
𝑞(𝑥)=𝑥
𝜇= 𝑒 𝑝 𝑥 .𝑑𝑥
𝜇= 𝑒 4 ln 𝑥
= 𝑥 4
𝑧= 1 𝜇 𝜇𝑞 𝑥 .𝑑𝑥
= 1 𝑥 𝑥 5 .𝑑𝑥
= 𝑥 𝐶 𝑥 4

8. Example 1 𝑧= 𝑥 2 6 + 𝐶 𝑥 4 𝑧= 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 𝑥 2 6 + 𝐶 𝑥 4
Notice we have two constants of integration, which is normal for 2nd order DEs. Normally, we would be given two conditions to use to find them.
𝑧= 𝑥 𝐶 𝑥 4
𝑧= 𝑑𝑦 𝑑𝑥
𝑑𝑦 𝑑𝑥 = 𝑥 𝐶 𝑥 4
𝑦= 𝑥 𝐶 𝑥 4 .𝑑𝑥
= 𝑥 − 𝐶 3 𝑥 3 +𝐷

9. Example 2 𝑦 2 𝑑 2 𝑦 𝑑 𝑥 2 − 𝑑𝑦 𝑑𝑥 3 =0 𝑧= 𝑑𝑦 𝑑𝑥 𝑑𝑧 𝑑𝑥 = 𝑑 2 𝑦 𝑑 𝑥 2
𝑦 2 𝑑 2 𝑦 𝑑 𝑥 2 − 𝑑𝑦 𝑑𝑥 3 =0
𝑧= 𝑑𝑦 𝑑𝑥
𝑑𝑧 𝑑𝑥 = 𝑑 2 𝑦 𝑑 𝑥 2
𝑑𝑧 𝑑𝑥 = 𝑑𝑧 𝑑𝑦 ⋅ 𝑑𝑦 𝑑𝑥
𝑦 2 𝑧 𝑑𝑧 𝑑𝑦 − 𝑧 3 =0
𝑑 2 𝑦 𝑑 𝑥 2 = 𝑑𝑧 𝑑𝑦 ⋅𝑧
𝑦 2 𝑑𝑧 𝑑𝑦 = 𝑧 2
1 𝑧 2 .𝑑𝑧 = 𝑦 2 .𝑑𝑦
− 1 𝑧 =− 1 𝑦 +𝐶
1 𝑧 = 1 𝑦 +𝐶

10. Example 2 1 𝑧 = 1 𝑦 +𝐶 𝑧= 𝑑𝑦 𝑑𝑥 𝑑𝑥 𝑑𝑦 = 1 𝑦 +𝐶 𝑥= 1 𝑦 +𝐶 .𝑑𝑦
This is as nice as this equation is going to get – we can’t write 𝑦=𝑓(𝑥)
1 𝑧 = 1 𝑦 +𝐶
𝑧= 𝑑𝑦 𝑑𝑥
𝑑𝑥 𝑑𝑦 = 1 𝑦 +𝐶
𝑥= 1 𝑦 +𝐶 .𝑑𝑦
= ln 𝑦 +𝐶𝑦+𝐷

11. Do Now Any Questions? Delta Workbook Nothing this time. Workbook
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12. This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Aaron Stockdill
2016