september 6 math 2306 sec. 51 fall 2019 - faculty...

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September 6 Math 2306 sec. 51 Fall 2019 Section 4: First Order Equations: Linear General Solution of First Order Linear ODE I Put the equation in standard form y 0 + P (x )y = f (x ), and correctly identify the function P (x ). I Obtain the integrating factor μ(x )= exp (R P (x ) dx ) . I Multiply both sides of the equation (in standard form) by the integrating factor μ. The left hand side will always collapse into the derivative of a product d dx [μ(x )y ]= μ(x )f (x ). I Integrate both sides, and solve for y . y (x )= 1 μ(x ) Z μ(x )f (x ) dx = e - R P(x ) dx Z e R P(x ) dx f (x ) dx + C September 4, 2019 1 / 37

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Page 1: September 6 Math 2306 sec. 51 Fall 2019 - Faculty Webfacultyweb.kennesaw.edu/lritter/Sept6_2306_51_F19f.pdf · September 6 Math 2306 sec. 51 Fall 2019 Section 4: First Order Equations:

September 6 Math 2306 sec. 51 Fall 2019

Section 4: First Order Equations: Linear

General Solution of First Order Linear ODEI Put the equation in standard form y ′ + P(x)y = f (x), and correctly

identify the function P(x).I Obtain the integrating factor µ(x) = exp

(∫P(x)dx

).

I Multiply both sides of the equation (in standard form) by theintegrating factor µ. The left hand side will always collapse intothe derivative of a product

ddx

[µ(x)y ] = µ(x)f (x).

I Integrate both sides, and solve for y .

y(x) =1

µ(x)

∫µ(x)f (x)dx = e−

∫P(x) dx

(∫e∫

P(x) dx f (x)dx + C)

September 4, 2019 1 / 37

Page 2: September 6 Math 2306 sec. 51 Fall 2019 - Faculty Webfacultyweb.kennesaw.edu/lritter/Sept6_2306_51_F19f.pdf · September 6 Math 2306 sec. 51 Fall 2019 Section 4: First Order Equations:

Solve the ODE

dydx

+y = 3xe−x

September 4, 2019 2 / 37

Page 3: September 6 Math 2306 sec. 51 Fall 2019 - Faculty Webfacultyweb.kennesaw.edu/lritter/Sept6_2306_51_F19f.pdf · September 6 Math 2306 sec. 51 Fall 2019 Section 4: First Order Equations:

September 4, 2019 3 / 37

Page 4: September 6 Math 2306 sec. 51 Fall 2019 - Faculty Webfacultyweb.kennesaw.edu/lritter/Sept6_2306_51_F19f.pdf · September 6 Math 2306 sec. 51 Fall 2019 Section 4: First Order Equations:

Solve the IVP

xdydx−y = 2x2, x > 0 y(1) = 5

September 4, 2019 4 / 37

Page 5: September 6 Math 2306 sec. 51 Fall 2019 - Faculty Webfacultyweb.kennesaw.edu/lritter/Sept6_2306_51_F19f.pdf · September 6 Math 2306 sec. 51 Fall 2019 Section 4: First Order Equations:

September 4, 2019 5 / 37

Page 6: September 6 Math 2306 sec. 51 Fall 2019 - Faculty Webfacultyweb.kennesaw.edu/lritter/Sept6_2306_51_F19f.pdf · September 6 Math 2306 sec. 51 Fall 2019 Section 4: First Order Equations:

September 4, 2019 6 / 37

Page 7: September 6 Math 2306 sec. 51 Fall 2019 - Faculty Webfacultyweb.kennesaw.edu/lritter/Sept6_2306_51_F19f.pdf · September 6 Math 2306 sec. 51 Fall 2019 Section 4: First Order Equations:

VerifyJust for giggles, lets verify that our solution y = 2x2 + 3x really doessolve the differential equation we started with

xdydx− y = 2x2.

September 4, 2019 7 / 37

Page 8: September 6 Math 2306 sec. 51 Fall 2019 - Faculty Webfacultyweb.kennesaw.edu/lritter/Sept6_2306_51_F19f.pdf · September 6 Math 2306 sec. 51 Fall 2019 Section 4: First Order Equations:

Steady and Transient States

For some linear equations, the term yc decays as x (or t) grows. Forexample

dydx

+ y = 3xe−x has solution y =32

x2e−x + Ce−x .

Here, yp =32

x2e−x and yc = Ce−x .

Such a decaying complementary solution is called a transient state.

The corresponding particular solution is called a steady state.

September 4, 2019 8 / 37