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85_pdfsam_math 54 differential equation solutions odd

# 85_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.6 11. From dx dy = xy y 2 x 2 = y x y x 2 we conclude that given equation is homogeneous. Let u = y/x . Then y = xu and y = u + xu . Substitution yields u + x du dx = u u 2 x du dx = u 2 du u 2 = dx x du u 2 = dx x 1 u = ln | x | + C x y = ln | x | + C y = x ln | x | + C . Note that, solving this equation, we have performed two divisions: by x 2 and u 2 . In doing this, we lost two solutions, x 0 and u 0. (The latter gives y 0.) Therefore, a general solution to the given equation is y = x ln | x | + C , x 0 , and y 0 . 13. Since we can express f ( t, x ) in the form G ( x/t ), that is, (dividing numerator and denominator by t 2 ) x 2 + t t 2 + x 2 tx = ( x/t ) 2 + ( x/t ) 2 ( x/t ) , the equation is homogeneous. Substituting v = x/t
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