96_pdfsam_math 54 differential equation solutions odd

96_pdfsam_math 54 differential equation solutions odd - ( x...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 2 7. This equation is separable. Separating variables and integrating, we get t 3 y 2 dt = t 4 y 6 dy dt t = dy y 8 ln | t | + C 1 = 1 7 y 7 y =(7ln | t | + C ) 1 / 7 . The function t 0 is also a solution. (We lost it when divided the equation by t 4 .) 9. The given diFerential equation can be written in the form dy dx + 1 3 x y = x 3 y 1 . This is a Bernoulli equation with n = 1, P ( x )=1 / (3 x ), and Q ( x )= x/ 3. To transform this equation into a linear equation, we ±rst multiply by y to obtain y dy dx + 1 3 x y 2 = 1 3 x. Next we make the substitution v = y 2 .S in c e v 0 =2 yy 0 , the transformed equation is 1 2 v 0 + 1 3 x v = 1 3 x, v 0 + 2 3 x v = 2 3 x. (2.23) The above equation is linear, so we can solve it for
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ( x ) is found to be ( x ) = exp Z 2 3 x dx = exp 2 3 ln | x | = x 2 / 3 . Multiplying equation (2.23) by x 2 / 3 gives x 2 / 3 v + 2 3 x 1 / 3 v = 2 3 x 5 / 3 ( x 2 / 3 v ) = 2 3 x 5 / 3 . We now integrate both sides and solve for v to nd x 2 / 3 v = Z 2 3 x 5 / 3 dx = 1 4 x 8 / 3 + C v = 1 4 x 2 + Cx 2 / 3 . Substituting v = y 2 gives the solution y 2 = 1 4 x 2 + Cx 2 / 3 ( x 2 + 4 y 2 ) x 2 / 3 = 4 C or, cubing both sides, ( x 2 + 4 y 2 ) 3 x 2 = C 1 , where C 1 := (4 C ) 3 is an arbitrary constant. 92...
View Full Document

This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

Ask a homework question - tutors are online