{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chap9_Sec5

# Chap9_Sec5 - 9 DIFFERENTIAL EQUATIONS DIFFERENTIAL...

This preview shows pages 1–14. Sign up to view the full content.

DIFFERENTIAL EQUATIONS DIFFERENTIAL EQUATIONS 9

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 52

Chap9_Sec5 - 9 DIFFERENTIAL EQUATIONS DIFFERENTIAL...

This preview shows document pages 1 - 14. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online