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Section 6.3 Day 1 Separation of Variable
AP Calculus AB
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Learning Targets Solve more challenging differential equations using separation of variables
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Example 1 Solve π₯ 2 +3π¦ ππ¦ ππ₯ =0 1. 3π¦ ππ¦ = β π₯ 2 ππ₯
π¦ ππ¦ = β π₯ 2 ππ₯ π¦ 2 =β 1 3 π₯ 3 +πΆ 3. π¦ 2 =β 2 9 π₯ 3 +πΆ
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Example 2 sin π¦ π¦ β² = cos π₯ sin π¦ ππ¦= cos π₯ ππ₯ β cos π¦ = sin π₯ +πΆ
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Example 3 Solve π π¦ π₯ π¦ β² π π¦ +1 =2 1. π π¦ π₯ π π¦ +1 ππ¦ ππ₯ =2
1. π π¦ π₯ π π¦ +1 ππ¦ ππ₯ =2 2. π π¦ π π¦ +1 ππ¦= 2 π₯ ππ₯ π π¦ π π¦ +1 ππ¦= 2 π₯ ππ₯ 4. ln π π¦ +1 =2 ln π₯ +πΆ
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Example 4 Solve π₯ 2 +4 ππ¦ ππ₯ =π₯π¦ 1. π₯ 2 +4 ππ¦=π₯π¦ππ₯
1. π₯ 2 +4 ππ¦=π₯π¦ππ₯ ππ¦ π¦ = π₯ π₯ 2 +4 ππ₯ 3. ln π¦ = 1 2 ln π₯ πΆ 4. π¦ =Β± π πΆ π₯ 2 +4 5. π¦=πΆ π₯ 2 +4
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Example 5 Given π¦ 0 =1, find the particular solution of the equation π₯π¦ππ₯+ π β π₯ 2 π¦ 2 β1 ππ¦=0 1. π β π₯ 2 π¦ 2 β1 ππ¦=βπ₯π¦ππ₯ 2. π¦ 2 β1 π¦ ππ¦=βπ₯ π π₯ 2 ππ₯ 3. π¦β 1 π¦ ππ¦ = βπ₯ π π₯ 2 ππ₯ π¦ 2 β ln π¦ =β 1 2 π π₯ 2 +πΆ 5. πΆ=1 6. π¦ 2 β ln π¦ π π₯ 2 =2
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Example 6 Find the equation of the curve that passes through the point (1,3) and has a slope of π¦ π₯ 2 at any point (π₯,π¦) 1. ππ¦ ππ₯ = π¦ π₯ 2 π¦ ππ¦ = π₯ 2 ππ₯ 3. ln π¦ =β 1 π₯ +πΆ 4. π¦=πΆ π β 1 π₯ , πΆ=3π 5. π¦=3 π β 1 π₯ +1
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