If $ f(x) $ satisfies the differential equation $ \frac{d y}{d x}=(x-y)^{2} $ and given that $ y(1)=1 $, then

Question

If $ f(x) $ satisfies the differential equation $ \frac{d y}{d x}=(x-y)^{2} $ and given that $ y(1)=1 $, then, $ -\ln \left|\frac{1-x+y}{1+x-y}\right|=2(x-1) $, $ \ln \left|\frac{2-y}{2-x}\right|=x+y-1 $, $ \ln \left|\frac{1-x+y}{1+x-y}\right|=2(x-1) $, $ \frac{1}{2} \ln \left|\frac{1-x+y}{1+x-y}\right|+\ln |x|=0 $

MARKS App · Accepted Answer

x − y = t 1 − d y d x = d t d x Hence 1 − d t d x = t 2 ⇒ 1 − t 2 = d t d x ⇒ ∫ d t 1 − t 2 = ∫ d x ⇒ 1 2 log 1 + t t − 1 = x + c 1 2 log x − y + 1 x − y − 1 = x + c At x = 1 and y = 1 , we get c = − 1 Hence 1 2 log x − y + 1 x − y − 1 = x − 1 Hence − ln 1 − x + y 1 + x − y = 2 x − 1

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