{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

nagle_differential_equations_ISM_Part15

# nagle_differential_equations_ISM_Part15 - Chapter 2 32 To...

This preview shows pages 1–2. Sign up to view the full content.

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.

Unformatted text preview: Chapter 2 32. To obtain a homogeneous equation, we make a substitution x = u + h , y = v + k with h and k satisfying 2 h + k + 4 = 0 h- 2 k- 2 = 0 ⇒ h =- 6 5 , k =- 8 5 . This substitution yields (2 u + v ) du + ( u- 2 v ) dv = 0 ⇒ dv du = v + 2 u 2 v- u = ( v/u ) + 2 2( v/u )- 1 . We now let z = v/u (so, v = z + uz ) and conclude that z + u dz du = z + 2 2 z- 1 ⇒ u dz du = z + 2 2 z- 1- z =- 2 z 2 + 2 z + 2 2 z- 1 ⇒ Z (2 z- 1) dz z 2- z- 1 =- 2 Z du u ⇒ ln z 2- z- 1 =- 2 ln | u | + C 1 ⇒ ln u 2 z 2- u 2 z- u 2 = C 1 ⇒ ln v 2- uv- u 2 = C 1 ⇒ ln y + 8 5 2- y + 8 5 x + 6 5- x + 6 5 2 = C 1 ⇒ (5 y + 8) 2- (5 y + 8) (5 x + 6)- (5 x + 6) 2 = C, where C = ± 25 e C 1 6 = 0 is any constant. Separating variables, we lost two constant solutions z ( u ), which are the zeros of the polynomial z 2- z- 1. They can be included in the above formula by taking C = 0. Therefore, a general solution is given by (5 y + 8) 2- (5 y + 8) (5 x + 6)- (5 x + 6) 2 = C, where C is an arbitrary constant. 34. In Problem 2, we found that the given equation can be written as a Bernoulli equation, dx dt- 1 2 t x =- t 2 x- 1 . Thus, 2 x dx dt- 1 t x 2 =- t ⇒ ( v = x 2 ) dv dt- 1 t v =- t....
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

nagle_differential_equations_ISM_Part15 - Chapter 2 32 To...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online