SATHEE: Differential Equations 1 Question 2

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) $\log_{e} | \frac{2 - y}{2 - x} | = 2 (y - 1)$

(b) $- \log_{e} | \frac{1 + x - y}{1 - x + y} | = x + y - 2$

(d) $- \log_{e} | \frac{1 - x + y}{1 + x - y} | = 2 (x - 1)$

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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 $x - y = t$

$\begin{aligned} \Rightarrow & 1 - \frac{d y}{d x} & = \frac{d t}{d x} \Rightarrow \frac{d y}{d x} = 1 - \frac{d t}{d x} \\ \Rightarrow & 1 - \frac{d t}{d x} & = t^{2} [∵ \frac{d y}{d x} = (x - y)^{2}] \\ \Rightarrow & \frac{d t}{d x} & = 1 - t^{2} \Rightarrow \int \frac{d t}{1 - t^{2}} = \int d x \end{aligned}$

[separating the variables]

$\begin{aligned} \Rightarrow \frac{1}{2} \log_{e} \frac{1 + t}{1 - t} = x + C \\ \int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} \log_{e} | \frac{a + x}{a - x} | + C \\ \Rightarrow \frac{1}{2} \log_{e} \frac{1 + x - y}{1 - x + y} = x + C [∵ t = x - y] \end{aligned}$

Since, $y = 1$ when $x = 1$ , therefore

$\begin{aligned} \frac{1}{2} \log_{e} \frac{1 + 0}{1 + 0} = 1 + C \\ \Rightarrow & C = - 1 \\ ∴ & \frac{1}{2} \log_{e} \frac{1 + x - y}{1 - x + y} = x - 1 \\ \Rightarrow & - \log_{e} | \frac{1 - x + y}{1 + x - y} | = 2 (x - 1) \end{aligned}$

$[∵ \log \frac{1}{x} = \log x^{- 1} = - \log x]$

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

Answer:

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Differential Equations 1 Question 2

2. The solution of the differential equation, dydx=(x−y)2, when y(1)=1, is

Answer:

Engineering Preparation

Medical Preparation

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Syllabus

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2. The solution of the differential equation, $\frac{d y}{d x} = (x - y)^{2}$ , when $y (1) = 1$ , is