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ICSE Class 10 Answered

maths
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Asked by shaimaobaid30420 | 03 Nov, 2023, 02:20: PM
answered-by-expert Expert Answer
begin mathsize 14px style x space left parenthesis 1 minus y right parenthesis space d x space minus space left parenthesis space 1 minus space y squared right parenthesis left parenthesis x minus 1 right parenthesis d y space equals space 0 end style
 
Above expression is written as
 
begin mathsize 14px style fraction numerator d y over denominator d x end fraction space equals space fraction numerator x space left parenthesis space 1 minus space y right parenthesis over denominator left parenthesis 1 minus y squared right parenthesis left parenthesis x minus 1 right parenthesis end fraction space equals space fraction numerator x over denominator left parenthesis x minus 1 right parenthesis end fraction cross times fraction numerator 1 over denominator left parenthesis 1 plus y right parenthesis end fraction end style
 
begin mathsize 14px style left parenthesis 1 plus y right parenthesis space d y space equals space fraction numerator x over denominator open parentheses x minus 1 close parentheses end fraction space d x end style
 
begin mathsize 14px style left parenthesis 1 plus y right parenthesis space d y space equals space fraction numerator left parenthesis x minus 1 right parenthesis plus 1 over denominator open parentheses x minus 1 close parentheses end fraction space d x space end style
 
begin mathsize 14px style left parenthesis 1 plus y right parenthesis space d y space equals space space d x space plus space fraction numerator d x over denominator open parentheses x minus 1 close parentheses end fraction end style
By integrating both sides , we get
 
 
begin mathsize 14px style 1 half open parentheses 1 plus y close parentheses squared space equals space x space plus space log left parenthesis x minus 1 right parenthesis space plus space C
open parentheses 1 plus y close parentheses squared space equals space 2 x space plus space 2 space log left parenthesis x minus 1 right parenthesis space equals space 2 x space plus space log left parenthesis x minus 1 right parenthesis squared space plus space C
left parenthesis 1 plus y right parenthesis space equals space square root of 2 x space plus space log left parenthesis x minus 1 right parenthesis squared plus C end root space

y space equals space square root of 2 x space plus space log left parenthesis x minus 1 right parenthesis squared plus C end root space minus space 1 space
end style
Where C is constant of integration
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begin mathsize 14px style fraction numerator d y over denominator d x end fraction equals space square root of fraction numerator left parenthesis 1 minus y squared right parenthesis over denominator left parenthesis 1 minus x squared right parenthesis end fraction end root end style
begin mathsize 14px style fraction numerator d y over denominator square root of left parenthesis 1 minus y squared right parenthesis end root end fraction space equals space fraction numerator d x over denominator square root of left parenthesis 1 minus x squared right parenthesis end root end fraction end style
 
Let us use the following integration for above expression
 
Then we get
 
begin mathsize 14px style sin to the power of negative 1 end exponent y space equals space sin to the power of negative 1 end exponent x space plus space C

y space equals space sin space left parenthesis space sin to the power of negative 1 end exponent x space plus space C space right parenthesis end style
 
Answered by Thiyagarajan K | 03 Nov, 2023, 03:37: PM
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