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Unformatted text preview: MATH 3103 Fall 2011 Differential Equations Homework Set 4 Some Solutions The following solutions are not necessarily complete, but should highlight the main points for most questions. If you have any further questions regarding the problems and solutions, feel free to discuss them on WebCT. Solutions marked by * are taken from the publisher’s Instructor’s Solutions Manual for the textbook by Boyce and DiPrima, 9th edition (Wiley), while solutions marked by † (which appear to have been typed with a typewriter...) are from the publisher’s Student Solutions Manual; in some cases I have added comments in [brackets]. Section 2.2: * Section 2.2 # 14 : * Section 2.2 # 22 : * Section 2.2 # 24 : • Section 2.2 # 30 : Following the steps outlined in the problem, we begin with dy dx = y 4 x x y = ( y/x ) 4 1 ( y/x ) . As indicated, the appropriate substitution for homogeneous equations is v = y/x . Hence y ( x ) = xv ( x ), so by the product rule, dy/dx = v + xdv/dx . Substituting, x dv dx + v = v 4 1 v = ⇒ x dv dx = v 4 1 v v = v 2 4 1 v , which is a separable equation. Separating (for v 2 6 = 4), we need to integrate (1 v ) / ( v 2 4), which we rewrite using partial fractions: 1 v v 2 4 = 1 v ( v + 2)( v 2) = A v + 2 + B v 2 = ( A + B ) v + 2( B A ) v 2 4 , which is true for all v (comparing coefficients of powers of v ) if A + B = 1 and 2( B A ) = 1, so A = 3 / 4, B = 1 / 4. Returning to the differential equation for v ( x ), we have Z 1 v v 2 4 dv = Z 1 4 3 v + 2 + 1 v 2 dv = Z 1 x dx + c = ln  x  + c, so 3ln  v + 2  + ln  v 2  = ln  v + 2  3  v 2  = 4(ln  x  + c ) = ln  x  4 4 c = ⇒  v + 2  3  v 2  = C  x  4 (where C = e 4 c > 0) ....
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This note was uploaded on 01/30/2012 for the course MATH 310 303 taught by Professor Rwk during the Spring '11 term at Simon Fraser.
 Spring '11
 RWK
 Math, Differential Equations, Equations

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