Differential Equations 1 Question 2
2. The solution of the differential equation, $\frac{d y}{d x}=(x-y)^{2}$, when $y(1)=1$, is
(2019 Main, 11 Jan II)
(a) $\log _e\left|\frac{2-y}{2-x}\right|=2(y-1)$
(b) $-\log _e\left|\frac{1+x-y}{1-x+y}\right|=x+y-2$
(c) $\log _e\left|\frac{2-x}{2-y}\right|=x-y$
(d) $-\log _e\left|\frac{1-x+y}{1+x-y}\right|=2(x-1)$
Show Answer
Answer:
Correct Answer: 2. (d)
Solution:
- We have, $\frac{d y}{d x}=(x-y)^{2}$ which is a differential equation of the form
$$ \frac{d y}{d x}=f(a x+b y+c) $$
Put $x-y=t$
$$ \begin{aligned} \Rightarrow & 1-\frac{d y}{d x} & =\frac{d t}{d x} \Rightarrow \frac{d y}{d x}=1-\frac{d t}{d x} \\ \Rightarrow & 1-\frac{d t}{d x} & =t^{2} \quad\left[\because \frac{d y}{d x}=(x-y)^{2}\right] \\ \Rightarrow & \frac{d t}{d x} & =1-t^{2} \Rightarrow \int \frac{d t}{1-t^{2}}=\int d x \end{aligned} $$
[separating the variables]
$$ \begin{aligned} & \Rightarrow \quad \frac{1}{2} \log _e \frac{1+t}{1-t}=x+C \\ & \int \frac{d x}{a^{2}-x^{2}}=\frac{1}{2 a} \log _e\left|\frac{a+x}{a-x}\right|+C \\ & \Rightarrow \quad \frac{1}{2} \log _e \frac{1+x-y}{1-x+y}=x+C \quad[\because t=x-y] \end{aligned} $$
Since, $y=1$ when $x=1$, therefore
$$ \begin{aligned} & \frac{1}{2} \log _e \frac{1+0}{1+0}=1+C \\ \Rightarrow & \quad C=-1 \\ \therefore \quad & \quad \frac{1}{2} \log _e \frac{1+x-y}{1-x+y}=x-1 \\ \Rightarrow \quad & -\log _e\left|\frac{1-x+y}{1+x-y}\right|=2(x-1) \end{aligned} $$
$\left[\because \log \frac{1}{x}=\log x^{-1}=-\log x\right]$
