The solution of the differential equation (1+y²) dx = (^{-1} y - x) dy is

Correct Answer is : xetan⁡−1y=(tan⁡−1y−1)etan⁡−1y+Cx e^{^{-1} y} = (^{-1} y - 1) e^{^{-1} y} + Cxetan−1y=(tan−1y−1)etan−1y+C

x e^{^{-1} y} = (1 - ^{-1} y) e^{^{-1} y} + C

x e^{^{-1} y} = (^{-1} y - 1) e^{^{-1} y} + C

x = ^{-1} y - 1 + C e^{-^{-1} y}

Correct Answer is : xetan⁡−1y=(tan⁡−1y−1)etan⁡−1y+Cx e^{^{-1} y} = (^{-1} y - 1) e^{^{-1} y} + Cxetan−1y=(tan−1y−1)etan−1y+C

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CGPET 2012 Solved Paper

Section: Mathematics

Question : 146 of 150

Marks: +1, -0

(1+y^2) dx = (\tan^{-1} y - x) dy

Solution:

Given, differential equation can be written as

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

\; \frac{dx}{dy} + \; \frac{1}{1+y^2} \cdot x = \; \frac{\tan^{-1} y}{1+y^2}

It is a linear differential equation of the form

\; \frac{dx}{dy} + P x = Q

where,

\; P = \; \frac{1}{1+y^2}

and

Q = \; \frac{\tan^{-1} y}{1+y^2}

\therefore

Integrating factor

= e^{\int P dy}

\; = e^{\int \; \frac{1}{1+y^2} dy}

\; = e^{\tan^{-1} y}

\therefore

Solution is

e^{\tan^{-1} y} \cdot x = \int \; \frac{e^{\tan^{-1} y} \tan^{-1} y}{1+y^2} dy

Put

\tan^{-1} y = t

in right hand side,

\; \Rightarrow \; \; \frac{1}{1+y^2} dy = dt

\; \therefore \; \; x e^{\tan^{-1} y} = \int \; \underset{\text{II I}}{e^t t} dt

\; = t e^t - \int 1 \cdot e^t dt + C

\; = t e^t - e^t + C

\; = \tan^{-1} y \cdot e^{\tan^{-1} y} - e^{\tan^{-1} y} + C

\; \Rightarrow \; \; x e^{\tan^{-1} y} = (\tan^{-1} y - 1) e^{\tan^{-1} y} + C

\;

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