MTH306_notes_all - MTH306 Differential Equations Recitation...

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A First Course in Differential Equations with Modeling Applications
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Chapter 1 / Exercise 40
A First Course in Differential Equations with Modeling Applications
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MTH306 Differential Equations Recitation Notes 2012.5.29-2012.6.27 Yin Su Section 1.6 Bernoulli equations Differential equations of form d y d x = F ( ax + b y + c ) . – Use substitution v = ax + b y + c . Homogeneous equations d y d x = F ( y x ) . – Use substitution v = y x . Bernoulli equations d y d x + P ( x ) y = Q ( x ) y n . n = 0: Standard linear first order equation. (Section 1.5) n = 1: d y d x = ( Q ( x ) - P ( x )) y . Separable equation. (Section 1.4) – all the other n : do the substitution v = y 1 - n . Steps: (1) Substitution: v = y 1 - n . Then dv d x = ( 1 - n ) y - n d y d x (2) Multiply the original equation by ( 1 - n ) y - n . We have dv d x + ( 1 - n ) P ( x ) | {z } new P ( x ) v = ( 1 - n ) Q ( x ) | {z } new Q ( x ) (3) Solve (2) for v ( x ) and get the implicit solution for y . Example. §2.6 # 7 Solve the equation x y 2 y 0 = x 3 + y 3 . x y 2 + y 3
(3) Solve this equation and get the solution.
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A First Course in Differential Equations with Modeling Applications
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Chapter 1 / Exercise 40
A First Course in Differential Equations with Modeling Applications
Zill
Expert Verified
Equation dv d x + ( - 3 x ) v = 3 x 2 . Then P ( x ) = - 3 x , Q ( x ) = 3 x 2 . (1) Integrating factor: ρ ( x ) = e R - 3 x d x = e - 3ln x = x - 3 = 1 x 3 (2) Multiply both sides by the integrating factor, we have 1 x 3 v ( x ) = Z 1 x 3 · 3 x 2 d x = Z 3 x d x = 3ln | x | + C (3) Solve for v : y 3 = v = x 3 ( 3ln | x | + C ) Section 2.2 Equilibrium solutions and stability. Autonomous equation: d x dt = f ( x ) . ( x is a function of t .) Equilibrium solutions: the solution of f ( x ) = 0. (Critical points) Stability: Suppose x = c is a critical point. - Stable: Solution curves approach x = c as t + . - Unstable: Solution curves diverge away from x = c as t + . Steps to find the equilibrium solutions and stability: (1) Find critical points of autonomous 1st order ODE; (2) Classify critical points by analyzing sign of f ( x ) around critical points. (Pick points around the critical points.) (3) Draw the phase diagram. Example. §2.2 # 1 d x dt = x - 4. Solution: f ( x ) = x - 4. This is an autonomous equation. (1) Critical points: f ( x ) = x - 4 = 0 x = 4. (2) Stability at x = 4: If x > 4 and x is close to 4, f ( x ) = x - 4 > 0. Then x ( t ) % ; (We can pick x = 5.) If x < 4 and x is close to 4, f ( x ) = x - 4 < 0. Then x ( t ) & . (We can pick x = 3.) Then x = 4 is unstable. (3) Phase diagram: x = 4 is unstable: Example. §2.2 # 6 d x dt = 9 - x 2 . Solution: f ( x ) = 9 - x 2 . This is an autonomous equation. (1) Critical points: f ( x ) = 9 - x 2 = 0 x = 3, - 3.
(2) Stability at x = 3: If x > 3 and x is close to 3, f ( x ) = 9 - x 2 < 0. Then x ( t ) & ; (We can pick x = 4.) If x < 3 and x is close to 3, f ( x ) = 9 - x 2 > 0. Then x ( t ) % . (We can pick x = 2.) Then x = 3 is stable. Stability at x = - 3: If x > - 3 and x is close to - 3, f ( x ) = 9 - x 2 > 0. Then x ( t ) % ; (We can pick x = - 2.) If x < - 3 and x is close to - 3, f ( x ) = 9 - x 2 < 0. Then x ( t ) & . (We can pick x = - 4.) Then x = - 3 is unstable. (3) Phase diagram: x = 3 is stable; x = - 3 is unstable.

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