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UPSC NDA I 2018 Mathematics Solved Paper

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Question : 82 of 120

Marks: +1, -0

What is the solution of (1 +2x) dy – (1 – 2y) d

x = 0

?

$x - y - 2xy = c$
$y - x - 2xy = c$
$y + x - 2xy = c$
$x + y + 2xy = c$

Solution:

Concept:

Separation of variables and integration to solve a first-order differential equation.

Explanation:

Given:

(1 + 2x) \, dy - (1 - 2y) \, dx = 0

.

Rearrange:

(1 + 2x) \, dy = (1 - 2y) \, dx

.

Divide both sides by

(1 + 2x)(1 - 2y)

:

\frac{dy}{1 - 2y} = \frac{dx}{1 + 2x}

.

Integrate each side:

\int \frac{dy}{1 - 2y} = \int \frac{dx}{1 + 2x}

.

Evaluate the integrals:

-\frac{1}{2} \ln|1 - 2y| = \frac{1}{2} \ln|1 + 2x| + C_1

.

Multiply both sides by

2

and combine logs:

\ln|1 + 2x| + \ln|1 - 2y| = \ln C_2

.

So

\ln|(1 + 2x)(1 - 2y)| = \ln C_2

, hence

(1 + 2x)(1 - 2y) = C_2

.

Expand:

1 - 2y + 2x - 4xy = C_2

.

Rearrange:

2x - 2y - 4xy = C_2 - 1

.

Divide by

2

:

x - y - 2xy = C

, where

C

is an arbitrary constant.

Answer:

x - y - 2xy = c

(Option A).

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