The differential equation for which \sqrt{1+y^2}=Cxe^{Tan^{-1}x} is the general solution, of

Correct Answer is : xy(1+x2)dy−(1+y2)(1+x+x2)dx=0

xy(1+x^2)dy-e^{Tan^{-1}x}(1+x+x^2)dx=0

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

(1+y^2)Tan^{-1}x\frac{dy}{dx}=\frac{1+x^2}{xy}

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

TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

Question : 78 of 160

Marks: +1, -0

The differential equation for which √1+y2=CxeTan−1x is the general solution, of

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