MTH306_notes_all - MTH306 Differential Equations Recitation...

• Notes
• mie920418
• 58
• 100% (2) 2 out of 2 people found this document helpful

This preview shows page 1 - 4 out of 58 pages.

We have textbook solutions for you!
The document you are viewing contains questions related to this textbook. The document you are viewing contains questions related to this textbook.
Chapter 1 / Exercise 40
A First Course in Differential Equations with Modeling Applications
Zill Expert Verified
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.
We have textbook solutions for you!
The document you are viewing contains questions related to this textbook. The document you are viewing contains questions related to this textbook.
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.
• • • 