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UPSC NDA I 2014 Mathematics Paper

For the next three (03) items that follow:

The general solution of the differential equation

(x^2+x+1)dy+(y^2+y+1)dx=0

(x+y+1)=A(1+Bx+Cy+Dxy)

where B, C and D are constants and A is parameter.

[NDA I 2014]

Question : 105 of 120

Marks: +1, -0

What is D equal to?

Solution:

Given differential equation is

(x^2+x+1)dy+(y^2+y+1)dx=0

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

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

\Rightarrow \frac{dx}{1+x+x^2}+\frac{dy}{1+y+y^2}=0

\Rightarrow \int \frac{dx}{\left(x+\frac{1}{2}\right)^2+\frac{3^2}{4}} + \int \frac{dy}{\left(y+\frac{1}{2}\right)^2}=0

\Rightarrow \int \frac{dx}{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{4}\right)^2} + \int \frac{dy}{\left(y+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=0

(on integrating)

\Rightarrow \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left( \frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left( \frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)

=\frac{2}{\sqrt{3}} \tan^{-1} C_1 \left( \because \int \frac{dx}{a^2+x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} \right)

\Rightarrow \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2x+1}{\sqrt{3}} \right) + \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2y+1}{\sqrt{3}} \right) = \frac{2}{\sqrt{3}} \tan^{-1} C_1

\Rightarrow \tan^{-1} \left( \frac{2x+1}{\sqrt{3}} \right) + \tan^{-1} \left( \frac{2y+1}{\sqrt{3}} \right) = \tan^{-1} C_1

\Rightarrow \tan^{-1} \left\{ \frac{ \frac{2x+1}{\sqrt{3}} + \tan^{-1} \left( \frac{2y+1}{\sqrt{3}} \right) }{ 1 - \left( \frac{2x+1}{\sqrt{3}} \right) \left( \frac{2y+1}{\sqrt{3}} \right) } = \tan^{-1} C_1 \right\}

\left[\because \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)\right]

\Rightarrow \frac{ \sqrt{3}(2x+1)+(2y+1) }{ 3 - (2x+1)(2y+1) } = C_1

\Rightarrow \frac{ \sqrt{3}(2x+2y+2) }{ 3 - (4xy+2y+2x+1) } = C_1

\Rightarrow \frac{ 2\sqrt{3}(x+y+1) }{ -4xy-2y-2x+2 } = C_1

\Rightarrow 2\sqrt{3}(x+y+1) = C_1 (2-2x-2y-4xy)

\Rightarrow 2\sqrt{3}(x+y+1) = 2 C_1 (1-x-y-2xy)

\Rightarrow (x+y+1) = \frac{C_1}{\sqrt{3}} (1-x-y-2xy)

on comparing with

(x+y+1)=A(1+Bx+Cy+Dxy)

Here A is parameter and B, C and D are constants

The value of

D=-2

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