102_pdfsam_math 54 differential equation solutions odd

# 102_pdfsam_math 54 differential equation solutions odd - x...

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Chapter 2 which is a function of x alone. Thus, the equation has a special integrating factor µ ( x )=exp ±Z 2 x dx ² = x 2 . Multiplying the equation by µ ( x ), we Fnd that F ( x, y )= Z x 2 ( 4 xy 3 9 y 2 +4 xy 2 ) dx = x 4 y 3 3 x 3 y 2 + x 4 y 2 + g ( y ) ∂F ∂y =3 x 4 y 2 6 x 3 y +2 x 4 y + g 0 ( y )= x 2 N ( x, y )= x 2 ( 3 x 2 y 2 6 xy +2 x 2 y ) g 0 ( y )=0 g ( y ) 0 F ( x, y )= x 4 y 3 3 x 3 y 2 + x 4 y 2 = C is a general solution. 31. In this problem, ∂M ∂y = 1 , ∂N ∂x =1 , and so ∂M/∂y ∂N/∂x N = 2 x . Therefore, the equation has a special integrating factor µ ( x )=exp ³Z ± 2 x ² dx ´ = x 2 . We multiply the given equation by µ ( x ) to get an exact equation. µ
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Unformatted text preview: x β y x 2 ΒΆ dx + 1 x dy = 0 β F ( x, y ) = Z Β± 1 x Β² dy = y x + h ( x ) β βF βx = β y x 2 + h ( x ) = x β y x 2 β h ( x ) = x β h ( x ) = x 2 2 , and a general solution is given by F ( x, y ) = y x + x 2 2 = C and x β‘ . (The latter has been lost in multiplication by Β΅ ( x ).) Substitution the initial values, y = 3 when x = 1, yields 3 1 + 1 2 2 = C β C = 7 2 . 98...
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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