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ENGR213_Week_2

# ENGR213_Week_2 - Slide 1 of 15 Plan for week 2 Concordia...

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Unformatted text preview: Slide 1 of 15 Plan for week 2 Concordia University - ENGR 213 Fall 2011 - class sections R and X Instructor: Eugene Kritchevski KEY CONCEPTS for WEEK 2 How to solve a first order DE? - Section 2.1 Solution curves without a solution (you are responsible for studying this section) 2.1.1 Direction fields 2.1.2 Autonomous DE - Section 2.2 Separable equations - Section 2.3 Linear equations 2 ENGR213 Week 2_cdf.cdf Slide 2 of 15 Section 2.1.1 Direction fields IDEA: The first order DE dy f x, y dx has a geometric meaning. At the point (x,y), the solution curve has slope equal to f(x,y). We know the tangent line at every point and we are trying to figure out what the curve is. ENGR213 Week 2_cdf.cdf Slide 3 of 15 Example f(x,y)=-2xy DE: y’ f(x,y) = Direction field 4 2 Out[22]= 0 2 4 4 2 0 2 4 We are trying to find curves with prescribed tangent lines 3 4 ENGR213 Week 2_cdf.cdf 4 2 0 Out[23]= 2 4 4 2 0 2 4 Solution curve to IVP y' f x,y , y x0 y0 y 4 2 y0 Out[33]= 0 x x0 2 4 4 2 0 2 4 After section 2.3 you will be able to find the explicit solution for such DE! ENGR213 Week 2_cdf.cdf Slide 4 of 15 Section 2.1.2 Autonomous First-Order DE A first-order DE of the form dy fy dx is called autonomous. Main feature: f does not depend on x direction field of DE does not depend on x Important property: If y=Φ(x) is a solution then, for any constant c, the function y=Φ(x-c) is also a solution. Proof: for any x we have Φ’x)=f(Φ(x)) ( so d dx Φx c Φ' x c fΦx c Equilibrium points: If f(y*)=0 then the constant function y=y* is a solution. Phase portrait (read the textbook!): - look at all the equlibrium points and the intervals in-between consecutive equilibrium points - the sign of f will tell you if a solution is increasing or decreasing 5 ENGR213 Week 2_cdf.cdf Slide 5 of 15 Example The logistic equation y’ y(1-y) = Direction field: 1.5 1.0 y 6 0.5 Out[26]= 0.0 0.5 4 2 0 2 x Discussion: - Hoes does the direction field change with x? - Does existence and uniqueness Theorem apply? - What are the equilibrium points? - Where are solutions increasing? - Where are solutions decreasing? - Sketch a phrase portrait - Can we sketch solution curves without solving the DE? - What are the attractors and the repellers? Interactive visualization of the solution curve to IVP 4 ENGR213 Week 2_cdf.cdf 1.5 y 1.0 Out[29]= 0.5 0.0 0.5 4 2 0 x 2 4 7 8 ENGR213 Week 2_cdf.cdf Slide 6 of 15 Section 2.2 What is a separable equation and how to solve it A first-order DE of the form dy hxgy dx is called separable. IDEA: we can treat the variables x and y separately to find a solution to the DE 1 dy h x dx gy 1 dy h x dx gy Gy Hx c where G’y)=1/g(y) and H’x)=h(x) and c is an arbitrary real constant ( ( We got a 1-parameter family of implicit solutions: Gy Hx c Differentiate implicitly (chain rule) with respect to x to double-check that DE is verified: Gy dy dx dy Hx G' y 1 dx g y c H' x hx ENGR213 Week 2_cdf.cdf dy hxgy dx SINGULAR SOLUTIONS: If g(y*)=0 then the constant function y=y* is a solution to the DE. 9 10 ENGR213 Week 2_cdf.cdf Slide 7 of 15 Example 1 Solve DE (1+x)dy - ydx = 0 Solution: the DE can be rewritten as dy 1 dx 1 y x so it is separable. dy dx y 1 1 x 1 dy y 1 dx x ln y ln 1 x y 1 x ec1 y ec1 1 x c1 Where c1 is an arbitrary real constant Relabel : c= ec1 and obtain the family of solutions y c1 x, c 0 One singular solution: y=0 so actually the family of solutions is y c1 x, c ENGR213 Week 2_cdf.cdf Slide 8 of 15 Example 2 Solve IVP dy x dx , y4 3 y Solution: The DE is separable ydy xdx ydy xdx y2 x2 2 2 C x2 y2 2C Initial condition y4 42 3 3 2 2C Implicit solution x2 y2 25 Explicit solution to IVP: y 25 x2 Are there any singular solutions that we missed? Why not y 25 x2 ? Direction field for the DE 25 2C 11 12 ENGR213 Week 2_cdf.cdf 10 5 Out[30]= 0 5 10 10 5 0 5 10 ENGR213 Week 2_cdf.cdf Slide 9 of 15 Example 3 Solve DE dy y2 4 dx Note that DE is autonomous (y’depends on y but not on x) EQUILIBRIUM SOLTIONS: y=2 and y=-2 FAMILY OF SOLUTIONS: dy y2 dx 4 dy y 2 dx y 2 In order to integrate the left hand side, use partial fractions 1 y 2 14 y y 2 14 y 2 2 So dy y 2 dx y 2 14 14 y y 2 1 y ln ln dy dx 2 1 dy 2 y y2 y2 ln y 4 dx 2 y y 2 y 2 4x 2 e4 xc 2 c 4x c 13 14 ENGR213 Week 2_cdf.cdf y 2 y case 1 : 2 e4 2 e4 y 2 y y y e4 xc 2 xc 2 e4 1 e4 e4 xc y e4 xc xc 2 e4 xc 2 xc 1 y xc ,c Discussion for case 1: - What is the geometric meaning of the parameter c ? - On which interval is the solution defined ? Solution curves for case 1: c 1 4 2 Out[31]= 4 2 2 4 2 4 y 2 y 2 y case 2 : 2 y e4 2 e4 xc xc y e4 xc 2 e4 xc ENGR213 Week 2_cdf.cdf y y e4 xc 2 e4 2 e4 e4 xc 2 xc 1 y 1 xc ,c Solution curves for case 2: c 1 4 2 Out[32]= 4 2 2 2 4 Discussion for case 2: - What is the geometric meaning of the parameter c ? - On which interval is the solution defined ? 4 15 16 ENGR213 Week 2_cdf.cdf Slide 10 of 15 Section 2.3 Linear (first order) equations a1 x dy a0 x y dx gx When g(x)=0, the DE is called homogeneous. When g(x) 0, the DE is called nonhomogeneous. Standard form: divide DE by a1 (x) dy a0 x dx gx a1 x y a1 x get dy DE Pxy fx dx where P(x)= a0 x a1 x and f(x)= gx a1 x We want solutions on an interval I where P and f are continuous. Homogeneous equation corresponding to (DE): DE hom dy Pxy 0 dx Fundametal connection: between (DE) and (DE hom): 1) If ynon hom is a solution to (DE) and yhom is a solution to (DE hom) then y yhom ynon hom is a solution to (DE) 2) If y1 and y2 are solutions to (DE), then y2 y1 is a solution to (DE hom) 3) If yH is the general solution to (DE hom) and ypart is a particular solution to (DE), then yH ypart is the general solution to (DE) ENGR213 Week 2_cdf.cdf Slide 11 of 15 How to solve a first-order linear DE DE dy Pxyx fx dx Solution uses an auxiliary function (called the integrating factor) x Ix Exp Pt t x0 (you can use any x0 you want Key properties of I(x): I' x I x0 any antiderivative of P(t) will do the job) PxIx 1 Multiply (DE) by I(x): DE Ix Ix dy dx dy IxPxyx I' x y x Ixfx Ixfx dx d Ixyx Ixfx dx x Ixyx I x0 y x0 Itft x0 x Ixyx y x0 Itft t Itft t x0 x Ixyx y x0 x0 1 yx x Ix y x0 Itft x0 1-pararameter family of solutions: yx 1 Ix x c Itft x0 t t t 17 18 ENGR213 Week 2_cdf.cdf If we impose the initial condition y x0 yx 1 Ix y0 , then the solution is x y0 Itft x0 t ENGR213 Week 2_cdf.cdf Slide 12 of 15 Example 1: y’ y=6 -3 19 20 ENGR213 Week 2_cdf.cdf Slide 13 of 15 Example 2: xy’ = x 6ex -4y ENGR213 Week 2_cdf.cdf Slide 14 of 15 Example 3: (x2-9)y’ =0 +xy 21 22 ENGR213 Week 2_cdf.cdf Slide 15 of 15 Example 4: y’ =x +y ...
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