Solve of the differential equation dy/dx = {1+y²{(^{-1}y)-x}

Correct Answer is : xetan⁡−1yxe^{^{-1}y}xetan−1y = etan⁡−1y((tan⁡−1y)−1)+ce^{^{-1}y}((^{-1}y)-1)+cetan−1y((tan−1y)−1)+c

xe^{^{-1}y} = e^{-^{-1}y}((^{-1}y)-1)+c

xe^{^{-1}y} = e^{^{-1}y}((^{-1}y)-1)+c

xe^{^{-1}y} = e^{^{-1}y}((^{-1}y)+1)+c

xe^{^{-1}y} = e^{-^{-1}y}((^{-1}y)+1)+c

Correct Answer is : xetan⁡−1yxe^{^{-1}y}xetan−1y = etan⁡−1y((tan⁡−1y)−1)+ce^{^{-1}y}((^{-1}y)-1)+cetan−1y((tan−1y)−1)+c

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AP EAMCET Engineering 21 Sep 2020 Shift 2 Paper

Section: Mathematics

Question : 78 of 160

Marks: +1, -0

\frac{dy}{dx}

\frac{1+y^2}{(\tan^{-1}y)-x}

Solution:

(b) Given differential equation

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

\Rightarrow \;\frac{dx}{dy}+\;\frac{x}{1+y^2}=\;\frac{\tan^{-1}y}{1+y^2} \;\text{ is a linear }\;

\;\text{ differential equation, so. }\;

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

So, the solution is

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

\Rightarrow x e^{\tan^{-1}y} = \int t \cdot e^t dt + c, \;\; \{\;\text{ where }\; t=\tan^{-1}y\}

\Rightarrow x e^{\tan^{-1}y} = t e^t - \int e^t + c

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

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

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