Differential Equations Ncert Solutions Maths class 12th Maths Class 12th Differential Equations

106. Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t)

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Vishal Baghel | Contributor-Level 10

4 months ago

$\begin{array}{l} (x - y) (d x + d y) = d x - d y \\ \Rightarrow (x - y + 1) d y = (1 - x + y) d x \\ \Rightarrow \frac{d y}{d x} = \frac{1 - x + y}{x - y + 1} \\ \Rightarrow \frac{d y}{d x} = \frac{1 - (x - y)}{1 + (x - y)} . . . . . . . . . . (1) \\ L e t, x - y = t \\ \Rightarrow \frac{d}{d x} (x - y) = \frac{d t}{d x} \\ \Rightarrow 1 - \frac{d y}{d x} = \frac{d t}{d x} \\ \Rightarrow 1 - \frac{d t}{d x} = \frac{d y}{d x} \end{array}$

Substituting the values of $x - y$ and $\frac{d y}{d x}$ in equation (1), we get:

$\begin{array}{l} 1 - \frac{d t}{d x} = \frac{1 - t}{1 + t} \\ \Rightarrow \frac{d t}{d x} = 1 - (\frac{1 - t}{1 + t}) \\ \Rightarrow \frac{d t}{d x} = \frac{(1 + t) - (1 - t)}{1 + t} \\ \Rightarrow \frac{d t}{d x} = \frac{2 t}{1 + t} \end{array}$

$\begin{array}{l} \Rightarrow (\frac{1 + t}{t} d t) = 2 d x \\ \Rightarrow (1 + \frac{1}{t}) d t = 2 d x . . . . . . . . . . (2) \end{array}$

Integrating both sides, we get:

$\begin{array}{l} t + l o g | t | = 2 x + C \\ \Rightarrow (x - y) + l o g | x - y | = 2 x + C \\ \Rightarrow l o g | x - y | = x + y + C . . . . . . . . . . (3) \end{array}$

$N o w, y = - 1, a t, x = 0$

Therefore, equation (3) becomes:

$l o g 1 = 0 - 1 + C$

$\Rightarrow C = 1$

Substituting $C = 1$ in equation (3), we get:

$o g | x - y | = x + y + 1$

This is the required particular solution of the given differential equation .

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106. Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t)

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