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

For each of the differential equations in Questions from 11 to 15, find the particular solution satisfying the given condition

70. (x + y) dy + (x – y) dx = 0; y = 1 when x = 1

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Vishal Baghel | Contributor-Level 10

4 months ago

The given D.E. is

$\begin{array}{l} (x + y) d y + (x - y) d x = 0 \\ = (x + y) d y = - (x - y) d x \\ = \frac{d y}{d x} = \frac{y - x}{x + y} = \frac{\frac{y - x}{x}}{\frac{x + y}{y}} = \frac{\frac{y}{x} - 1}{1 + \frac{y}{x}} = f (\frac{y}{x}) \end{array}$

i.e, homogenous

Let, $y = v x = v = \frac{y}{x}$ so that $\frac{d y}{d x} = v + y \frac{d y}{d x}$ in the D.E.

Then, $v + x \frac{d v}{d x} = \frac{v - 1}{v + 1}$

$\begin{array}{l} = x \frac{d v}{d x} = \frac{v - 1}{v + 1} - v = \frac{v - 1 - v^{2} - v}{v + 1} = - \frac{(v^{2} + 1)}{v + 1} \\ = [\frac{v + 1}{v^{2} + 1}] d v = - \frac{d x}{x} \end{array}$

Integrating both sides,

$\begin{array}{l} \int \frac{v + 1}{v^{2} + 1} d v = \int - \frac{d x}{x} \\ = \frac{1}{2} \int \frac{2 v}{v^{2} + 1} d v + \int \frac{1}{v^{2} + 1} d v = - l o g x + c \\ = \frac{l o g | v^{2} + 1 |}{2} + t a n^{- 1} v = - l o g x + c \end{array}$

Putting back $v = \frac{y}{x}$ we get,

$\begin{array}{l} = \frac{1}{2} l o g | \frac{y^{2}}{x^{2}} + 1 | + t a n^{- 1} \frac{y}{x} = - l o g x + c \\ = \frac{1}{2} [l o g (y^{2} + x^{2}) - l o g x^{2}] + t a n^{- 1} \frac{y}{x} + l o g x = c \\ = \frac{1}{2} l o g (y^{2} + x^{2}) - \frac{1}{2} l o g x^{2} + l o g x + t a n^{- 1} \frac{y}{x} = c \\ = \frac{1}{2} l o g (y^{2} + x^{2}) - l o g x + l o g x + t a n^{- 1} \frac{y}{x} = c \\ = \frac{1}{2} l o g (y^{2} + x^{2}) + t a n^{- 1} \frac{y}{x} = c \end{array}$

Given, $y = 1, w h e n, x = 1$

So, $= \frac{1}{2} l o g (1^{2} + 1^{2}) + t a n^{- 1} \frac{1}{1} = c$

$= \frac{1}{2} l o g 2 + \frac{π}{4} = c$

Hence, the particular solution is

$\frac{1}{2} l o g (1^{2} + 1^{2}) + t a n^{- 1} \frac{y}{x} = \frac{1}{2} l o g 2 + \frac{π}{4}$

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For each of the differential equations in Questions from 11 to 15, find the particular solution satisfying the given condition

70. (x + y) dy + (x – y) dx = 0; y = 1 when x = 1

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For each of the differential equations in Questions from 11 to 15, find the particular solution satisfying the given condition 70. (x + y) dy + (x – y) dx = 0; y = 1 when x = 1

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For each of the differential equations in Questions from 11 to 15, find the particular solution satisfying the given condition

70. (x + y) dy + (x – y) dx = 0; y = 1 when x = 1