Section 6.3 Day 1 Separation of Variable AP Calculus AB

Learning Targets Solve more challenging differential equations using separation of variables

Example 1 Solve 𝑥 2 +3𝑦 𝑑𝑦 𝑑𝑥 =0 1. 3𝑦 𝑑𝑦 = − 𝑥 2 𝑑𝑥 1. 3𝑦 𝑑𝑦 = − 𝑥 2 𝑑𝑥 2. 3 2 𝑦 2 =− 1 3 𝑥 3 +𝐶 3. 𝑦 2 =− 2 9 𝑥 3 +𝐶

Example 2 sin 𝑦 𝑦 ′ = cos 𝑥 sin 𝑦 𝑑𝑦= cos 𝑥 𝑑𝑥 − cos 𝑦 = sin 𝑥 +𝐶

Example 3 Solve 𝑒 𝑦 𝑥 𝑦 ′ 𝑒 𝑦 +1 =2 1. 𝑒 𝑦 𝑥 𝑒 𝑦 +1 𝑑𝑦 𝑑𝑥 =2 1. 𝑒 𝑦 𝑥 𝑒 𝑦 +1 𝑑𝑦 𝑑𝑥 =2 2. 𝑒 𝑦 𝑒 𝑦 +1 𝑑𝑦= 2 𝑥 𝑑𝑥 3. 𝑒 𝑦 𝑒 𝑦 +1 𝑑𝑦= 2 𝑥 𝑑𝑥 4. ln 𝑒 𝑦 +1 =2 ln 𝑥 +𝐶

Example 4 Solve 𝑥 2 +4 𝑑𝑦 𝑑𝑥 =𝑥𝑦 1. 𝑥 2 +4 𝑑𝑦=𝑥𝑦𝑑𝑥 1. 𝑥 2 +4 𝑑𝑦=𝑥𝑦𝑑𝑥 2. 𝑑𝑦 𝑦 = 𝑥 𝑥 2 +4 𝑑𝑥 3. ln 𝑦 = 1 2 ln 𝑥 2 +4 +𝐶 4. 𝑦 =± 𝑒 𝐶 𝑥 2 +4 5. 𝑦=𝐶 𝑥 2 +4

Example 5 Given 𝑦 0 =1, find the particular solution of the equation 𝑥𝑦𝑑𝑥+ 𝑒 − 𝑥 2 𝑦 2 −1 𝑑𝑦=0 1. 𝑒 − 𝑥 2 𝑦 2 −1 𝑑𝑦=−𝑥𝑦𝑑𝑥 2. 𝑦 2 −1 𝑦 𝑑𝑦=−𝑥 𝑒 𝑥 2 𝑑𝑥 3. 𝑦− 1 𝑦 𝑑𝑦 = −𝑥 𝑒 𝑥 2 𝑑𝑥 4. 1 2 𝑦 2 − ln 𝑦 =− 1 2 𝑒 𝑥 2 +𝐶 5. 𝐶=1 6. 𝑦 2 − ln 𝑦 2 + 𝑒 𝑥 2 =2

Example 6 Find the equation of the curve that passes through the point (1,3) and has a slope of 𝑦 𝑥 2 at any point (𝑥,𝑦) 1. 𝑑𝑦 𝑑𝑥 = 𝑦 𝑥 2 2. 1 𝑦 𝑑𝑦 = 1 𝑥 2 𝑑𝑥 3. ln 𝑦 =− 1 𝑥 +𝐶 4. 𝑦=𝐶 𝑒 − 1 𝑥 , 𝐶=3𝑒 5. 𝑦=3 𝑒 − 1 𝑥 +1