The solution of the differential equation, `dy/dx=(x-y)^(2)`, when `y(1)=1,` is

Options

A. `log_(e) abs((2-y)/(2-x))=2(y-1)`
B. `-log_(e)abs((1+x-y)/(1-x+y))=x+y-2`
C. `log_(e) abs((2-x)/(2-y))=x-y`
D. `-log_(e) abs((1-x+y)/(1+x-y))=2(x-1)`

Answer

D

Explanation

We have `dy/dx = (x-y)^(2)` which is a differential equation of
the form
`dy/dx =f(ax+by+c)`
Put `x-y=t`
`rArr 1-dy/dx=(dt)/dx rArr dy/dx =1-(dt)/dx`
`rArr 1-(dt)/dx=t^(2) [therefore dy/dx= (x-y)^(2)]`
`rArr (dt)/dx=1-t^(2) rArr int (dt)/(1-t^(2))=int dx`
[separating the variables]
`rArr 1/2log_(e)((1+t)/(1-t))=x+C`
`[int dx/(a^(2)-x^(2))=1/(2a)log_(e) abs((a+x)/(a-x))+C]`
`rArr 1/2log_(e)((1+x-y)/(1-x+y))=x+C [therefore t=x-y]`
Since, y=1 when x=1, therefore
`1/2log_(e)((1+0)/(1+0))=1+C`
`rArr C=-1 [therefore log1=0]`
`therefore 1/2log_(e)((1+x-y)/(1-x+y))=x+1`
`rArr -log_(e)abs((1-x+y)/(1+x-y))=2(x-1)`
`[therefore log frac{1}x=logx^(-1)=-log x]`