Chap9_Sec5.ppt - 9 DIFFERENTIAL EQUATIONS DIFFERENTIAL EQUATIONS 9.5 Linear Equations In this section we will learn How to solve linear equations using

# Chap9_Sec5.ppt - 9 DIFFERENTIAL EQUATIONS DIFFERENTIAL...

• 52

This preview shows page 1 - 14 out of 52 pages.

DIFFERENTIAL EQUATIONS DIFFERENTIAL EQUATIONS 9
9.5 Linear Equations DIFFERENTIAL EQUATIONS In this section, we will learn: How to solve linear equations using an integrating factor.
LINEAR EQUATIONS A first-order linear differential equation is one that can be put into the form where P and Q are continuous functions on a given interval. This type of equation occurs frequently in various sciences, as we will see. ( ) ( ) dy P x y Q x dx Equation 1
LINEAR EQUATIONS An example of a linear equation is xy’ + y = 2 x because, for x ≠ 0, it can be written in the form 1 ' 2 y y x Equation 2
LINEAR EQUATIONS Notice that this differential equation is not separable. It’s impossible to factor the expression for y’ as a function of x times a function of y .
LINEAR EQUATIONS However, we can still solve the equation by noticing, by the Product Rule, that xy’ + y = ( xy ) So, we can rewrite the equation as: ( xy ) = 2 x
LINEAR EQUATIONS If we now integrate both sides, we get: xy = x 2 + C or y = x + C / x If the differential equation had been in the form of Equation 2, we would have had to initially multiply each side of the equation by x .
INTEGRATING FACTOR It turns out that every first-order linear differential equation can be solved in a similar fashion by multiplying both sides of Equation 1 by a suitable function I ( x ). This is called an integrating factor.
LINEAR EQUATIONS We try to find I so that the left side of Equation 1, when multiplied by I ( x ), becomes the derivative of the product I ( x ) y : I ( x )( y’ + P ( x ) y ) = ( I ( x ) y ) Equation 3
LINEAR EQUATIONS If we can find such a function I , then Equation 1 becomes: ( I ( x ) y ) = I ( x ) Q ( x ) Integrating both sides, we would have: I ( x ) y = I ( x ) Q ( x ) dx + C
LINEAR EQUATIONS So, the solution would be: 1 ( ) ( ) ( ) ( ) y x I x Q x dx C I x Equation 4
LINEAR EQUATIONS To find such an I , we expand Equation 3 and cancel terms: I ( x ) y’ + I ( x ) P ( x ) y = ( I ( x ) y ) = I ( x ) y + I ( x ) y’ I ( x ) P ( x ) = I ( x )
SEPERABLE DIFFERENTIAL EQUATIONS This is a separable differential equation for I , which we solve as follows: where A = ± e C .

#### You've reached the end of your free preview.

Want to read all 52 pages?

• Fall '10
• capretta

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern