83_pdfsam_math 54 differential equation solutions odd

# 83_pdfsam_math 54 differential equation solutions odd - G (...

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Exercises 2.6 ∂F ∂y = ∂y [( y x +1) e x ]= ( y x +1) ∂y e x = e x . Now we can use the result of Problem 32 to derive an equation for the orthogonal trajectories (i.e., velocity potentials) of the given family of curves: ∂F ∂y dx ∂F ∂x dy =0 e x dx ( y x ) e x dy =0 dx +( x y ) dy =0 . (b) In the diFerential equation dx +( x y ) dy =0 , M =1and N = x y . Therefore, ∂N/∂x ∂M/∂y M = ( x y ) /∂x (1) /∂y (1) =1 , and an integrating factor µ ( y )isg ivenby µ ( y )=exp ±R (1) dy ² = e y . Multiplying the equation from part (a) by µ ( y ) yields an exact equation, and we look for its solutions of the form G ( x, y )= c . e y dx +( x y ) e y dy =0 G ( x, y )= Z e y dx = xe y + g ( y ) ∂G ∂y = xe y + g 0 ( y )=( x y ) e y g 0 ( y )= ye y g ( y )= Z ( ye y ) dy = ³ ye y Z e y dy ´ = ye y + e y
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Unformatted text preview: G ( x, y ) = xe y ye y + e y = c or x = y 1 + ce y . EXERCISES 2.6: Substitutions and Transformations, page 78 1. We can write the equation in the form dy dx = ( y 4 x 1) 2 = [( y 4 x ) 1] 2 = G ( y 4 x ) , where G ( t ) = ( t 1) 2 . Thus, it is of the form dy/dx = G ( ax + by ). 3. In this equation, the variables are x and t . Its coecients, t + x + 2 and 3 t x 6, are linear functions of x and t . Therefore, given equation is an equation with linear coecients. 79...
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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 University of California, Berkeley.

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