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homework set 2-Solutions - BILKENT UNIVERSITY Department of...

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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 x dx + 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
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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 0 = 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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