85_pdfsam_math 54 differential equation solutions odd

85_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 2.6 11. From dx xy − y 2 y2 y = =− dy x2 x x we conclude that given equation is homogeneous. Let u = y/x. Then y = xu and y = u + xu . Substitution yields u+x ⇒ ⇒ du du = u − u2 = −u2 ⇒ x ⇒ dx dx dx du 1 = − ⇒ = ln |x| + C 2 u x u x x = ln |x| + C ⇒ y= . y ln |x| + C − du dx = 2 u x Note that, solving this equation, we have performed two divisions: by x2 and u2 . 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 t2 ) √ (x/t)2 + (x/t)2 x2 + t t2 + x2 = , tx (x/t) the equation is homogeneous. Substituting v = x/t and dx/dt = v + tdv/dt into the equation yields √ √ dv dv 1 + v2 1 + v2 v+t =v+ ⇒ t = . dt v dt v This transformed equation is separable. Thus we have √ 1 v dv = dt 2 t 1+v ⇒ √ 1 + v 2 = ln |t| + C, where we have integrated with the integration on the left hand side being accomplished by the substitution u = 1 + v 2 . Substituting x/t for v in this equation gives the solution to the original equation which is 1+ x2 = ln |t| + C. t2 81 ...
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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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