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CBSE Class 12-science Answered

S o l v e space t h e space f o l l o w i n g space d i f f e r e n t i a l space e q u a t i o n colon
x squared left parenthesis y plus 1 right parenthesis d x plus y squared left parenthesis x minus 1 right parenthesis d y equals 0Please provide solution to this question
Asked by Topperlearning User | 04 Nov, 2016, 08:40: PM
answered-by-expert Expert Answer

x squared left parenthesis y plus 1 right parenthesis d x plus y squared left parenthesis x minus 1 right parenthesis d y equals 0
rightwards double arrow x squared left parenthesis y plus 1 right parenthesis d x equals negative y squared left parenthesis x minus 1 right parenthesis d y
rightwards double arrow fraction numerator x squared d x over denominator left parenthesis x minus 1 right parenthesis end fraction equals negative fraction numerator y squared d y over denominator left parenthesis y plus 1 right parenthesis end fraction
integral fraction numerator x squared d x over denominator left parenthesis x minus 1 right parenthesis end fraction equals negative integral fraction numerator y squared d y over denominator left parenthesis y plus 1 right parenthesis end fraction
rightwards double arrow integral fraction numerator x squared minus 1 plus 1 over denominator left parenthesis x minus 1 right parenthesis end fraction d x equals negative integral fraction numerator y squared minus 1 plus 1 over denominator left parenthesis y plus 1 right parenthesis end fraction d y
rightwards double arrow integral fraction numerator x squared minus 1 over denominator left parenthesis x minus 1 right parenthesis end fraction d x plus integral fraction numerator 1 over denominator x minus 1 end fraction d x equals negative integral fraction numerator y squared minus 1 over denominator left parenthesis y plus 1 right parenthesis end fraction d y minus integral fraction numerator 1 over denominator y plus 1 end fraction d y
rightwards double arrow integral left parenthesis x plus 1 right parenthesis d x plus integral fraction numerator 1 over denominator x minus 1 end fraction d x equals negative integral left parenthesis y minus 1 right parenthesis d y minus integral fraction numerator 1 over denominator y plus 1 end fraction d y
rightwards double arrow fraction numerator left parenthesis x plus 1 right parenthesis squared over denominator 2 end fraction plus log open vertical bar x minus 1 close vertical bar equals negative fraction numerator left parenthesis y minus 1 right parenthesis squared over denominator 2 end fraction minus log open vertical bar y plus 1 close vertical bar plus C
rightwards double arrow fraction numerator left parenthesis x plus 1 right parenthesis squared over denominator 2 end fraction plus fraction numerator left parenthesis y minus 1 right parenthesis squared over denominator 2 end fraction plus log open vertical bar x minus 1 close vertical bar plus log open vertical bar y plus 1 close vertical bar equals C
rightwards double arrow x squared plus 1 plus 2 x plus y squared plus 1 minus 2 y plus 2 log open vertical bar left parenthesis x minus 1 right parenthesis left parenthesis y plus 1 right parenthesis close vertical bar equals 2 C
rightwards double arrow x squared plus 2 left parenthesis x minus y right parenthesis plus y squared plus 2 log open vertical bar left parenthesis x minus 1 right parenthesis left parenthesis y plus 1 right parenthesis close vertical bar equals 2 C minus 2 equals C to the power of apostrophe space i s space t h e space r e q u i r e d space d i f f e r e n t i a l
e q u a t i o n

Answered by | 04 Nov, 2016, 10:40: PM
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