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Section 6.3 Day 1 Separation of Variable

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Presentation on theme: "Section 6.3 Day 1 Separation of Variable"β€” Presentation transcript:

1 Section 6.3 Day 1 Separation of Variable
AP Calculus AB

2 Learning Targets Solve more challenging differential equations using separation of variables

3 Example 1 Solve π‘₯ 2 +3𝑦 𝑑𝑦 𝑑π‘₯ =0 1. 3𝑦 𝑑𝑦 = βˆ’ π‘₯ 2 𝑑π‘₯
𝑦 𝑑𝑦 = βˆ’ π‘₯ 2 𝑑π‘₯ 𝑦 2 =βˆ’ 1 3 π‘₯ 3 +𝐢 3. 𝑦 2 =βˆ’ 2 9 π‘₯ 3 +𝐢

4 Example 2 sin 𝑦 𝑦 β€² = cos π‘₯ sin 𝑦 𝑑𝑦= cos π‘₯ 𝑑π‘₯ βˆ’ cos 𝑦 = sin π‘₯ +𝐢

5 Example 3 Solve 𝑒 𝑦 π‘₯ 𝑦 β€² 𝑒 𝑦 +1 =2 1. 𝑒 𝑦 π‘₯ 𝑒 𝑦 +1 𝑑𝑦 𝑑π‘₯ =2
1. 𝑒 𝑦 π‘₯ 𝑒 𝑦 +1 𝑑𝑦 𝑑π‘₯ =2 2. 𝑒 𝑦 𝑒 𝑦 +1 𝑑𝑦= 2 π‘₯ 𝑑π‘₯ 𝑒 𝑦 𝑒 𝑦 +1 𝑑𝑦= 2 π‘₯ 𝑑π‘₯ 4. ln 𝑒 𝑦 +1 =2 ln π‘₯ +𝐢

6 Example 4 Solve π‘₯ 2 +4 𝑑𝑦 𝑑π‘₯ =π‘₯𝑦 1. π‘₯ 2 +4 𝑑𝑦=π‘₯𝑦𝑑π‘₯
1. π‘₯ 2 +4 𝑑𝑦=π‘₯𝑦𝑑π‘₯ 𝑑𝑦 𝑦 = π‘₯ π‘₯ 2 +4 𝑑π‘₯ 3. ln 𝑦 = 1 2 ln π‘₯ 𝐢 4. 𝑦 =Β± 𝑒 𝐢 π‘₯ 2 +4 5. 𝑦=𝐢 π‘₯ 2 +4

7 Example 5 Given 𝑦 0 =1, find the particular solution of the equation π‘₯𝑦𝑑π‘₯+ 𝑒 βˆ’ π‘₯ 2 𝑦 2 βˆ’1 𝑑𝑦=0 1. 𝑒 βˆ’ π‘₯ 2 𝑦 2 βˆ’1 𝑑𝑦=βˆ’π‘₯𝑦𝑑π‘₯ 2. 𝑦 2 βˆ’1 𝑦 𝑑𝑦=βˆ’π‘₯ 𝑒 π‘₯ 2 𝑑π‘₯ 3. π‘¦βˆ’ 1 𝑦 𝑑𝑦 = βˆ’π‘₯ 𝑒 π‘₯ 2 𝑑π‘₯ 𝑦 2 βˆ’ ln 𝑦 =βˆ’ 1 2 𝑒 π‘₯ 2 +𝐢 5. 𝐢=1 6. 𝑦 2 βˆ’ ln 𝑦 𝑒 π‘₯ 2 =2

8 Example 6 Find the equation of the curve that passes through the point (1,3) and has a slope of 𝑦 π‘₯ 2 at any point (π‘₯,𝑦) 1. 𝑑𝑦 𝑑π‘₯ = 𝑦 π‘₯ 2 𝑦 𝑑𝑦 = π‘₯ 2 𝑑π‘₯ 3. ln 𝑦 =βˆ’ 1 π‘₯ +𝐢 4. 𝑦=𝐢 𝑒 βˆ’ 1 π‘₯ , 𝐢=3𝑒 5. 𝑦=3 𝑒 βˆ’ 1 π‘₯ +1


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