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Published byAnn-Marie HΓ₯kansson Modified over 5 years ago
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Differential Equations: Separation of Variables
Silent Teacher Intelligent Practice Narration Your Turn ππ¦ ππ₯ =π₯π¦ π¦ ππ¦ ππ₯ =π₯ ππ¦ ππ₯ = 1 π₯π¦ Practice
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ππ¦ ππ₯ = π₯ 2 π¦ ππ¦ ππ₯ = π¦ 2 π₯ 1 π¦ 2 ππ¦ = π₯ ππ₯ β 1 π¦ = π₯ 2 2 +π
Worked Example Your Turn Find a general solution to Find a general solution to ππ¦ ππ₯ = π₯ 2 π¦ ππ¦ ππ₯ = π¦ 2 π₯ 1 π¦ 2 ππ¦ = π₯ ππ₯ β 1 π¦ = π₯ π π¦= 2 π΄β π₯ 2 where π΄ is a constant
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7. ππ¦ ππ₯ =2π₯π¦ 1.π¦ ππ¦ ππ₯ =2π₯ 8. ππ¦ ππ₯ =2π₯ π¦ 2 2. 2π₯ ππ¦ ππ₯ =π¦
Find the general solution to each differential equation in the form π=π(π) 7. ππ¦ ππ₯ =2π₯π¦ 8. ππ¦ ππ₯ =2π₯ π¦ 2 9. π₯ 2 ππ¦ ππ₯ =2 π¦ π₯ ππ¦ ππ₯ = π¦ 2 11. π¦ 2 ππ¦ ππ₯ = π₯ 12. ππ¦ ππ₯ = π₯π¦ 1.π¦ ππ¦ ππ₯ =2π₯ 2. 2π₯ ππ¦ ππ₯ =π¦ 3. 2 π₯ ππ¦ ππ₯ = 1 π¦ 4. 2 π₯ ππ¦ ππ₯ =π¦ 5. 2 π₯ ππ₯ ππ¦ =π¦ 6. 1 π₯π¦ ππ¦ ππ₯ = 1 2
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1. sin 2 π¦ ππ¦ ππ₯ = cos π₯ 7. ππ¦ ππ₯ =2π₯ π π¦ 2. sec 2 π¦ ππ¦ ππ₯ = cosec π₯
Find the general solution to each differential equation 1. sin 2 π¦ ππ¦ ππ₯ = cos π₯ 2. sec 2 π¦ ππ¦ ππ₯ = cosec π₯ 3. sec π¦ ππ¦ ππ₯ =2 cosec 2 π₯ 4. cosec π₯ ππ¦ ππ₯ =2 sec π¦ 5. ππ¦ ππ₯ =2 sec π₯ tan π¦ 6. ππ¦ ππ₯ =2 sec π¦ tan π₯ 7. ππ¦ ππ₯ =2π₯ π π¦ 8. ππ¦ ππ₯ =2π₯ π π₯+π¦ 9. ππ¦ ππ₯ = 2 π¦ π π₯+π¦ 10. ππ¦ ππ₯ = ln 2π₯ π¦ 11. ππ¦ ππ₯ = π₯ 2ln π¦ 12. ππ¦ ππ₯ = 2 ln π¦
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7. ππ¦ ππ₯ =2π₯π¦ 1.π¦ ππ¦ ππ₯ =2π₯ 8. ππ¦ ππ₯ =2π₯ π¦ 2 2. 2π₯ ππ¦ ππ₯ =π¦
Find the general solution to each differential equation in the form π=π(π) 7. ππ¦ ππ₯ =2π₯π¦ 8. ππ¦ ππ₯ =2π₯ π¦ 2 9. π₯ 2 ππ¦ ππ₯ =2 π¦ π₯ ππ¦ ππ₯ = π¦ 2 11. π¦ 2 ππ¦ ππ₯ = π₯ 12. ππ¦ ππ₯ = π₯π¦ 1.π¦ ππ¦ ππ₯ =2π₯ 2. 2π₯ ππ¦ ππ₯ =π¦ 3. 2 π₯ ππ¦ ππ₯ = 1 π¦ 4. 2 π₯ ππ¦ ππ₯ =π¦ 5. 2 π₯ ππ₯ ππ¦ =π¦ 6. 1 π₯π¦ ππ¦ ππ₯ = 1 2 π¦ 2 =2 π₯ 2 +π΄ π¦=Β± 2π₯ 2 +π΄ ln π¦ = π₯ 2 +π π¦=π΄ π π₯ 2 ln π¦ = 1 2 ln π₯ + ln π΄ π¦=π΄ π₯ β 1 π¦ = π₯ 2 +π π¦= 1 π΄β π₯ 2 π¦ 2 = π₯ π΄ π¦=Β± π₯ π΄ π¦ =β 1 π₯ +π π¦= 1 π₯ 2 β 2π π₯ + π 2 β 1 π¦ = π₯ +π π¦= 1 π΄β π₯ ln π¦ = π₯ π π¦=π΄ π π₯ 2 `4 π¦ 3 3 = π₯ 3 +π π¦= 2 π₯ 3 +π΄ 1 3 π¦=Β±2 ln π΄π₯ ln π¦ = π₯ π π¦=π΄ π π₯ 2 `4 2 π¦ = π₯ 3 +π π¦= π₯ 3 +π΄ 2 9
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1. sin 2 π¦ ππ¦ ππ₯ = cos π₯ 7. ππ¦ ππ₯ =2π₯ π π¦ 2. sec 2 π¦ ππ¦ ππ₯ = cosec π₯
Find the general solution to each differential equation β 1 π π¦ = π₯ 2 +π π¦= ln 1 π΄β π₯ 2 1. sin 2 π¦ ππ¦ ππ₯ = cos π₯ 2. sec 2 π¦ ππ¦ ππ₯ = cosec π₯ 3. sec π¦ ππ¦ ππ₯ =2 cosec 2 π₯ 4. cosec π₯ ππ¦ ππ₯ =2 sec π¦ 5. ππ¦ ππ₯ =2 sec π₯ tan π¦ 6. ππ¦ ππ₯ =2 sec π¦ tan π₯ 7. ππ¦ ππ₯ =2π₯ π π¦ 8. ππ¦ ππ₯ =2π₯ π π₯+π¦ 9. ππ¦ ππ₯ = 2 π¦ π π₯+π¦ 10. ππ¦ ππ₯ = ln 2π₯ π¦ 11. ππ¦ ππ₯ = π₯ 2ln π¦ 12. ππ¦ ππ₯ = 2 ln π¦ π¦ 2 β 1 2 sin 2π¦ = sin π₯ +π β 1 π π¦ =2π₯ π π₯ β2 π π₯ +π π π¦ = 1 2 π π₯ β2π₯ π π₯ +π΄ tan π¦ = ln π΄ β ln | cosec π₯+ cot π₯| tan π¦= ln π΄ cosec π₯+ cot π₯ βπ¦ π βπ¦ β π βπ¦ =2 π π₯ +π π βπ¦ π¦+1 =2 π π₯ +π ln sec π¦ + tan π¦ =β 2cot π₯ +π π¦ 2 2 =π₯ ln 2π₯+π₯+π π¦=Β± 2π₯ ln 2π₯ +2π₯+π΄ sin π¦=β2 cos π₯ +π π¦ ln π¦+π¦ = π₯ π ln sin π¦ = ln sec π¦ + tan π¦ + ln π΄ ln sin π¦ = ln (π΄( sec π¦+ tan π¦ ) ) π¦ ln π¦+π¦ =2π₯+π sin π¦ = ln sec π₯ + ln π΄ sin π¦ = ln (π΄ sec π₯ )
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