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Unformatted text preview: B ˙ ILKENT UNIVERSITY Department of Mathematics MATH 225, LINEAR ALGEBRA and DIFFERENTIAL EQUATIONS, Solution of Homework set 1 # 2 U. Mu˘gan June 13, 2008 Homework problems from the 2 nd Edition, SECTION 1.4 3 (3) ) 2 The given D.E. is separable, so if we separate the variables and integrate both sides, we obtain Z dy y = Z sin xdx + c, c = constant Integration yields ln y = cos x + c If we take the exponential of both sides we get the following general solution: y ( x ) = e cos x + c 6 (6) ) Z dy √ y = Z 3 √ xdx 2 √ y = 2 x 3 / 2 + c, c = constant y = φ ( x ) = ( x 3 / 2 + C ) 2 , C = constant . 9 (9) ) Similar to previous problem, D.E is separable: Z dy y = Z 2 1 x 2 dx If we use the partial fraction decomposition, we obtain Z dy y = Z • 1 1 + x + 1 1 x ‚ dx Hence, ln y = ln(1 x ) ln(1 + x ) + ln c Solving for y gives y ( x ) = c 1 + x 1 x . 1 I made every effort to avoid the calculation errors and/or typos while I prepared the solution set. You are responsible to check all the solutions and correct the errors if there is any. If you find any errors and/or misprints, please notify me. 2 The number in the parenthesis denotes the problem number in the International Edition 10 (10) ) Z dy (1 + y ) 2 = Z dx (1 + x ) 2 Integration yields, 1 1 + y = 1 1 + x c = 1 + c (1 + x ) 1 + x , c = constant 15 (15) ) Z " 2 y 2 1 y 4 # dy = Z • 1 x 1 x 2 ‚ dx. Hence the implicit solution is 2 y + 1 3 y 2 = ln  x  + 1 x + c, c = constant . 18 (18) ) Write the given D.E. in form of x 2 y = 1 x 2 + y 2 x 2 y 2 = (1 x 2 )(1 + y 2 ) then it is separable and hence Z dy 1 + y 2 = Z • 1 x 2 1 ‚ dx tan 1 y = 1 x x + c, c = constant ....
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This note was uploaded on 06/18/2010 for the course COMPUTER S Math 225 taught by Professor Yosumhoca during the Spring '10 term at Bilkent University.
 Spring '10
 YosumHoca

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