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Q.
The solution of the differential equation, $\frac{d y}{d x}=\frac{y-x}{y-x-1}$, given $y(-5)=-5$ represents
Differential Equations
Solution:
Answer is $(y-x)^2-2 y=10$
Put $y-x=t \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}+1$;
hence $\frac{ dt }{ dx }+1=\frac{ t }{ t -1} \Rightarrow \frac{ dt }{ dx }=\frac{ t }{ t -1}-1=\frac{ t - t +1}{ t -1} $
$\frac{ dt }{ dx }=\frac{1}{ t -1} ; \frac{( t -1)^2}{2}= x + C$
$( y - x -1)^2=2 x + C ; C =9$