98_pdfsam_math 54 differential equation solutions odd

# 98_pdfsam_math 54 differential equation solutions odd - −...

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Chapter 2 Thus we get z +3= ( x + C ) 2 4 . Finally, replacing z by 2 x y yields 2 x y +3= ( x + C ) 2 4 . Solving for y ,weobta in y =2 x +3 ( x + C ) 2 4 . 17. This equation is a Bernoulli equation with n = 2. So, we divide it by y 2 and substitute u = y 1 to get du +2 u =1 du 2 u = 1 µ ( θ )=exp ±Z ( 2) ² = e 2 θ d ( e 2 θ u ) = e 2 θ e 2 θ u = Z ( e 2 θ ) = e 2 θ 2 + C 1 y 1 = 1 2 + C 1 e 2 θ = 1+ Ce 2 θ 2 y = 2 1+ Ce 2 θ . This formula, together with y 0, gives a general solution to the given equation. 19. In the di±erential equation M ( x, y )= x 2 3 y 2 and N ( x, y )=2 xy . The di±erential equation is not exact because ∂M ∂y = 6 y 6 =2 x = ∂N ∂x . However, because ( ∂M/∂y ∂N/∂x ) /N =( 8 y ) / (2 xy
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Unformatted text preview: − 4 /x depends only on x , we can determine µ ( x ) from equation (8) on page 70 of the text. This gives µ ( x ) = exp ³Z − 4 x dx ´ = x − 4 . When we multiply the di±erential equation by µ ( x ) = x − 4 we get the exact equation ( x − 2 − 3 x − 4 y 2 ) dx + 2 x − 3 y dy = 0 . To ²nd F ( x, y ) we integrate ( x − 2 − 3 x − 4 y 2 ) with respect to x : F ( x, y ) = Z ( x − 2 − 3 x − 4 y 2 ) dx = − x − 1 + x − 3 y 2 + g ( y ) . 94...
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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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