Review_for_Exam__3-Answers

Review_for_Exam__3-Answers - ANSWERS to LBS 119 Practice...

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ANSWERS to LBS 119 – Practice problems for Exam # 3. 1. 4 2 ( ) 0 y x y dx xdy - + = First, since it is in this form, check to see if it is EXACT -- You should be able to see quickly that this one is not! Therefore, it must be either Separable, Linear, or Bernoulli. Divide through by dx 4 2 4 2 3 2 and get the dy/dx symbol isolated on one side. Now, get by itself on one side Notice that the y's will not factor out and separate from the x's, so it . , ( ) 0 + - + = = - + = - dy dx dy dx dy dy y so dx dx x y x y x x y x y x y 3 2 is not separable. This equation has both a first power of y and a higher power of y, therefore it is a Bernoulli equation, we want the single power of y on the side with dy/dx Bernoulli : 1 + = dy dx x y x y 1-n 1-2 -1 2 -2 2 1 3 3 standard form with n=2. dv dy Let v = y y = y and dx dx So, we should multiply the standard form by -y and then substitute. Now this is a LINEAR equation in s = 1 1 - - - = - - - - = - = - y dy y dx x dv v dx x y x x [ ] 1 1 - ln| | ln| | 1 x 1 2 2 2 tandard form. =e Multiply the standard form through by . The left side always becomes [ ]'. Substitute back into the equation and integrate both 1 1 μ - - - - = = = - = - = - dx x x v e e x or x dv x v x dx x v x 1 2 3 1 4 4 -1 1 4 Solve for y sides. Finish solving for v. Plug y back in. y = y = 3 3 3 3 - - - = - = - + = - + - + - + x v dx x dx x x v C x v Cx x Cx x Cx
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(1 ) 0; (0) 1 y dx x dy y + + = = First, since it is in this form, check to see if it is EXACT -- You should be able to see quickly that this one is not! Therefore, it must be either Separable, Linear, or Bernoulli. Divide through by dx and get the dy/dx symbol isolated on one side. Get the on one side by itself. Notice that this is separable, since we can factor the y's out from the x's! Get t (1 ) 0 (1 ) 1 + + = + = - - = + dy dx dy y x dx dy x y dx y dy dx x 1 2 1 2 1 2 1 2 he y's on the left and x's on the right. Integrate both sides and then solve for y. Use initial condi 1 1 1 1 1 1 1 ln ln |1 | 1 2 2 ln |1 | 1 ln |1 | 2 - = + - - = + = + = + - = + + - = + + - = + + 1 dy dx x y y dy dx x u x du dx du u u C y x C y x C y x C 1 2 2 tions to find C. 1
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This test prep was uploaded on 04/10/2008 for the course LBS 119 taught by Professor Hanninichols during the Spring '08 term at Michigan State University.

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Review_for_Exam__3-Answers - ANSWERS to LBS 119 Practice...

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