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83_pdfsam_math 54 differential equation solutions odd

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

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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 differential equation dx + ( x y ) dy = 0, M = 1 and N = x y . Therefore, ∂N/∂x ∂M/∂y M = ( x y ) /∂x (1) /∂y (1) = 1 , and an integrating factor µ ( y ) is given by µ ( y ) = exp (1) dy = 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 ) = e y dx = xe y + g ( y ) ∂G ∂y = xe y + g ( y ) = ( x y ) e y g ( y ) = ye y g ( y ) = ( ye y ) dy = ye y e y dy = ye y + e y . Thus, the velocity potentials are given by
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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 coeﬃcients, t + x + 2 and 3 t − x − 6, are linear functions of x and t . Therefore, given equation is an equation with linear coeﬃcients. 79...
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