Differential Equations Substitution for First and Second ORDER

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.

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.

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 2 + 4 𝑥 2 𝑥 2 + 𝑦 2 𝑥 2 =0 𝑑𝑦 𝑑𝑥 ⋅ 𝑦 𝑥 +4+ 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

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 2 +4= 𝑘 𝑥 4 −𝑧 2 𝑧 2 +4 .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 2 +4 = ln 𝑥 +𝐶

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

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 2 +4 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 4 𝑥 5 .𝑑𝑥 = 𝑥 2 6 + 𝐶 𝑥 4

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 2 6 + 𝐶 𝑥 4 𝑧= 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 𝑥 2 6 + 𝐶 𝑥 4 𝑦= 𝑥 2 6 + 𝐶 𝑥 4 .𝑑𝑥 = 𝑥 3 18 − 𝐶 3 𝑥 3 +𝐷

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 = 1 𝑦 2 .𝑑𝑦 − 1 𝑧 =− 1 𝑦 +𝐶 1 𝑧 = 1 𝑦 +𝐶

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

Do Now Any Questions? Delta Workbook Nothing this time. Workbook Pages 194-197, 201-203

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. Aaron Stockdill 2016