The solution of the differential equation (1+y²)+(x-e^{^{-1} y})dy/dx=0 is

Correct Answer is : 2xetan⁡−1y=e2tan⁡−1y+k2x e^{^{-1} y}=e²^{-1} y}+k2xetan−1y=e2tan−1y+k

Correct Answer is : 2xetan⁡−1y=e2tan⁡−1y+k2x e^{^{-1} y}=e²^{-1} y}+k2xetan−1y=e2tan−1y+k

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CGPET 2019 Solved Paper

Section: Mathematics

Question : 117 of 150

Marks: +1, -0

(1+y^{2})+(x-e^{\tan^{-1} y})\frac{dy}{dx}=0

Solution:

(1+y^{2})+(x-e^{\tan^{-1} y})\frac{dy}{dx}=0

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

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

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

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

.......(i)

This is first order linear diff. equation in

x

\text{I F}=\int\frac{1}{1+y^{2}}dy=e^{\tan^{-1} y}

The general solution of Eq. (i)

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

Put

\tan^{-1} y=t

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

\therefore x\cdot e^{t}=\int e^{t}\cdot e^{t}dt+c

\Rightarrow x\cdot e^{t}=\int e^{2t}dt+c

\Rightarrow x\cdot e^{t}=\frac{e^{2t}}{2}+c

Put the values of

t

\Rightarrow x\cdot e^{\tan^{-1} y}

=\frac{e^{2\tan^{-1} y}}{2}+c

\Rightarrow 2x e^{\tan^{-1} y}

=e^{2\tan^{-1} y}+2c

\Rightarrow 2x e^{\tan^{-1} y}=e^{2\tan^{-1} y}+k

[

Let

k=2c]

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