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Differential Equations: Separation of Variables

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Presentation on theme: "Differential Equations: Separation of Variables"β€” Presentation transcript:

1 Differential Equations: Separation of Variables
Silent Teacher Intelligent Practice Narration Your Turn 𝑑𝑦 𝑑π‘₯ =π‘₯𝑦 𝑦 𝑑𝑦 𝑑π‘₯ =π‘₯ 𝑑𝑦 𝑑π‘₯ = 1 π‘₯𝑦 Practice

2 𝑑𝑦 𝑑π‘₯ = π‘₯ 2 𝑦 𝑑𝑦 𝑑π‘₯ = 𝑦 2 π‘₯ 1 𝑦 2 𝑑𝑦 = π‘₯ 𝑑π‘₯ βˆ’ 1 𝑦 = π‘₯ 2 2 +𝑐
Worked Example Your Turn Find a general solution to Find a general solution to 𝑑𝑦 𝑑π‘₯ = π‘₯ 2 𝑦 𝑑𝑦 𝑑π‘₯ = 𝑦 2 π‘₯ 1 𝑦 2 𝑑𝑦 = π‘₯ 𝑑π‘₯ βˆ’ 1 𝑦 = π‘₯ 𝑐 𝑦= 2 π΄βˆ’ π‘₯ 2 where 𝐴 is a constant

3 7. 𝑑𝑦 𝑑π‘₯ =2π‘₯𝑦 1.𝑦 𝑑𝑦 𝑑π‘₯ =2π‘₯ 8. 𝑑𝑦 𝑑π‘₯ =2π‘₯ 𝑦 2 2. 2π‘₯ 𝑑𝑦 𝑑π‘₯ =𝑦
Find the general solution to each differential equation in the form π’š=𝒇(𝒙) 7. 𝑑𝑦 𝑑π‘₯ =2π‘₯𝑦 8. 𝑑𝑦 𝑑π‘₯ =2π‘₯ 𝑦 2 9. π‘₯ 2 𝑑𝑦 𝑑π‘₯ =2 𝑦 π‘₯ 𝑑𝑦 𝑑π‘₯ = 𝑦 2 11. 𝑦 2 𝑑𝑦 𝑑π‘₯ = π‘₯ 12. 𝑑𝑦 𝑑π‘₯ = π‘₯𝑦 1.𝑦 𝑑𝑦 𝑑π‘₯ =2π‘₯ 2. 2π‘₯ 𝑑𝑦 𝑑π‘₯ =𝑦 3. 2 π‘₯ 𝑑𝑦 𝑑π‘₯ = 1 𝑦 4. 2 π‘₯ 𝑑𝑦 𝑑π‘₯ =𝑦 5. 2 π‘₯ 𝑑π‘₯ 𝑑𝑦 =𝑦 6. 1 π‘₯𝑦 𝑑𝑦 𝑑π‘₯ = 1 2

4 1. sin 2 𝑦 𝑑𝑦 𝑑π‘₯ = cos π‘₯ 7. 𝑑𝑦 𝑑π‘₯ =2π‘₯ 𝑒 𝑦 2. sec 2 𝑦 𝑑𝑦 𝑑π‘₯ = cosec π‘₯
Find the general solution to each differential equation 1. sin 2 𝑦 𝑑𝑦 𝑑π‘₯ = cos π‘₯ 2. sec 2 𝑦 𝑑𝑦 𝑑π‘₯ = cosec π‘₯ 3. sec 𝑦 𝑑𝑦 𝑑π‘₯ =2 cosec 2 π‘₯ 4. cosec π‘₯ 𝑑𝑦 𝑑π‘₯ =2 sec 𝑦 5. 𝑑𝑦 𝑑π‘₯ =2 sec π‘₯ tan 𝑦 6. 𝑑𝑦 𝑑π‘₯ =2 sec 𝑦 tan π‘₯ 7. 𝑑𝑦 𝑑π‘₯ =2π‘₯ 𝑒 𝑦 8. 𝑑𝑦 𝑑π‘₯ =2π‘₯ 𝑒 π‘₯+𝑦 9. 𝑑𝑦 𝑑π‘₯ = 2 𝑦 𝑒 π‘₯+𝑦 10. 𝑑𝑦 𝑑π‘₯ = ln 2π‘₯ 𝑦 11. 𝑑𝑦 𝑑π‘₯ = π‘₯ 2ln 𝑦 12. 𝑑𝑦 𝑑π‘₯ = 2 ln 𝑦

5 7. 𝑑𝑦 𝑑π‘₯ =2π‘₯𝑦 1.𝑦 𝑑𝑦 𝑑π‘₯ =2π‘₯ 8. 𝑑𝑦 𝑑π‘₯ =2π‘₯ 𝑦 2 2. 2π‘₯ 𝑑𝑦 𝑑π‘₯ =𝑦
Find the general solution to each differential equation in the form π’š=𝒇(𝒙) 7. 𝑑𝑦 𝑑π‘₯ =2π‘₯𝑦 8. 𝑑𝑦 𝑑π‘₯ =2π‘₯ 𝑦 2 9. π‘₯ 2 𝑑𝑦 𝑑π‘₯ =2 𝑦 π‘₯ 𝑑𝑦 𝑑π‘₯ = 𝑦 2 11. 𝑦 2 𝑑𝑦 𝑑π‘₯ = π‘₯ 12. 𝑑𝑦 𝑑π‘₯ = π‘₯𝑦 1.𝑦 𝑑𝑦 𝑑π‘₯ =2π‘₯ 2. 2π‘₯ 𝑑𝑦 𝑑π‘₯ =𝑦 3. 2 π‘₯ 𝑑𝑦 𝑑π‘₯ = 1 𝑦 4. 2 π‘₯ 𝑑𝑦 𝑑π‘₯ =𝑦 5. 2 π‘₯ 𝑑π‘₯ 𝑑𝑦 =𝑦 6. 1 π‘₯𝑦 𝑑𝑦 𝑑π‘₯ = 1 2 𝑦 2 =2 π‘₯ 2 +𝐴 𝑦=Β± 2π‘₯ 2 +𝐴 ln 𝑦 = π‘₯ 2 +𝑐 𝑦=𝐴 𝑒 π‘₯ 2 ln 𝑦 = 1 2 ln π‘₯ + ln 𝐴 𝑦=𝐴 π‘₯ βˆ’ 1 𝑦 = π‘₯ 2 +𝑐 𝑦= 1 π΄βˆ’ π‘₯ 2 𝑦 2 = π‘₯ 𝐴 𝑦=Β± π‘₯ 𝐴 𝑦 =βˆ’ 1 π‘₯ +𝑐 𝑦= 1 π‘₯ 2 βˆ’ 2𝑐 π‘₯ + 𝑐 2 βˆ’ 1 𝑦 = π‘₯ +𝑐 𝑦= 1 π΄βˆ’ π‘₯ ln 𝑦 = π‘₯ 𝑐 𝑦=𝐴 𝑒 π‘₯ 2 `4 𝑦 3 3 = π‘₯ 3 +𝑐 𝑦= 2 π‘₯ 3 +𝐴 1 3 𝑦=Β±2 ln 𝐴π‘₯ ln 𝑦 = π‘₯ 𝑐 𝑦=𝐴 𝑒 π‘₯ 2 `4 2 𝑦 = π‘₯ 3 +𝑐 𝑦= π‘₯ 3 +𝐴 2 9

6 1. sin 2 𝑦 𝑑𝑦 𝑑π‘₯ = cos π‘₯ 7. 𝑑𝑦 𝑑π‘₯ =2π‘₯ 𝑒 𝑦 2. sec 2 𝑦 𝑑𝑦 𝑑π‘₯ = cosec π‘₯
Find the general solution to each differential equation βˆ’ 1 𝑒 𝑦 = π‘₯ 2 +𝑐 𝑦= ln 1 π΄βˆ’ π‘₯ 2 1. sin 2 𝑦 𝑑𝑦 𝑑π‘₯ = cos π‘₯ 2. sec 2 𝑦 𝑑𝑦 𝑑π‘₯ = cosec π‘₯ 3. sec 𝑦 𝑑𝑦 𝑑π‘₯ =2 cosec 2 π‘₯ 4. cosec π‘₯ 𝑑𝑦 𝑑π‘₯ =2 sec 𝑦 5. 𝑑𝑦 𝑑π‘₯ =2 sec π‘₯ tan 𝑦 6. 𝑑𝑦 𝑑π‘₯ =2 sec 𝑦 tan π‘₯ 7. 𝑑𝑦 𝑑π‘₯ =2π‘₯ 𝑒 𝑦 8. 𝑑𝑦 𝑑π‘₯ =2π‘₯ 𝑒 π‘₯+𝑦 9. 𝑑𝑦 𝑑π‘₯ = 2 𝑦 𝑒 π‘₯+𝑦 10. 𝑑𝑦 𝑑π‘₯ = ln 2π‘₯ 𝑦 11. 𝑑𝑦 𝑑π‘₯ = π‘₯ 2ln 𝑦 12. 𝑑𝑦 𝑑π‘₯ = 2 ln 𝑦 𝑦 2 βˆ’ 1 2 sin 2𝑦 = sin π‘₯ +𝑐 βˆ’ 1 𝑒 𝑦 =2π‘₯ 𝑒 π‘₯ βˆ’2 𝑒 π‘₯ +𝑐 𝑒 𝑦 = 1 2 𝑒 π‘₯ βˆ’2π‘₯ 𝑒 π‘₯ +𝐴 tan 𝑦 = ln 𝐴 βˆ’ ln | cosec π‘₯+ cot π‘₯| tan 𝑦= ln 𝐴 cosec π‘₯+ cot π‘₯ βˆ’π‘¦ 𝑒 βˆ’π‘¦ βˆ’ 𝑒 βˆ’π‘¦ =2 𝑒 π‘₯ +𝑐 𝑒 βˆ’π‘¦ 𝑦+1 =2 𝑒 π‘₯ +𝑐 ln sec 𝑦 + tan 𝑦 =βˆ’ 2cot π‘₯ +𝑐 𝑦 2 2 =π‘₯ ln 2π‘₯+π‘₯+𝑐 𝑦=Β± 2π‘₯ ln 2π‘₯ +2π‘₯+𝐴 sin 𝑦=βˆ’2 cos π‘₯ +𝑐 𝑦 ln 𝑦+𝑦 = π‘₯ 𝑐 ln sin 𝑦 = ln sec 𝑦 + tan 𝑦 + ln 𝐴 ln sin 𝑦 = ln (𝐴( sec 𝑦+ tan 𝑦 ) ) 𝑦 ln 𝑦+𝑦 =2π‘₯+𝑐 sin 𝑦 = ln sec π‘₯ + ln 𝐴 sin 𝑦 = ln (𝐴 sec π‘₯ )


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