JEE PYQ: Differential Equation Question 47
Question 47 - 2019 (11 Jan Shift 2)
The solution of the differential equation $\frac{dy}{dx} = (x-y)^2$, when $y(1) = 1$, is:
(1) $\log_e\left|\frac{2-x}{2-y}\right| = x - y$
(2) $-\log_e\left|\frac{1-x+y}{1+x-y}\right| = 2(x-1)$
(3) $-\log_e\left|\frac{1+x-y}{1-x+y}\right| = x + y - 2$
(4) $\log_e\left|\frac{2-y}{2-x}\right| = 2(y-1)$
Show Answer
Answer: (2)
Solution
Let $t = x - y$: $\frac{dt}{dx} = 1 - t^2$. $\frac{dt}{1-t^2} = dx$. $\frac{1}{2}\ln\left|\frac{1+t}{1-t}\right| = x + c$… wait, $\frac{1}{2}\ln\left|\frac{t-1}{t+1}\right| = -x + c$. At $(1,1)$, $t = 0$: $c = -1$. So $-\frac{1}{2}\ln\left|\frac{1-t}{1+t}\right| = -(x-1)$. Substituting $t = x - y$: $-\log_e\left|\frac{1-x+y}{1+x-y}\right| = 2(x-1)$.
BETA
