The general solution of the differential equation ((1+x^2)y'+2xy/2x)=1+x^2, is (Consider _e x= x) (where C is an arbitrary constant) Option 1: y=^-1(1+x^2)-1/1+x^2(1+x^2)+C Option 2: y=^-1(1+x^2)(1+x^2)+C Option 3: y=(1+x^2)^-1(1+x^2)-(1+x^2)^2/2+C Option 4: y=^-1(1+x^2)-1/2(1+x^2)|x^4+2x^2+2|+C

The answer is 'y=^-1(1+x^2)-1/2(1+x^2)|x^4+2x^2+2|+C'. View full solution:

CUET Mathematics - The general solution of the differential equation ((1+x^2)y'+2xy/2x)=1+x^2, is (Consider _e x= x) (where C is an arbitrary constant) | PYQs + Solutions

CUET Mathematics 2026 18th May Shift 2

Calculus

Hard

core

$y=\tan^{-1}(1+x^2)-\dfrac{1}{1+x^2}\log(1+x^2)+C$

$y=\tan^{-1}(1+x^2)(1+x^2)+C$

$y=(1+x^2)\tan^{-1}(1+x^2)-\dfrac{(1+x^2)^2}{2}+C$

$y=\tan^{-1}(1+x^2)-\dfrac{1}{2(1+x^2)}\log|x^4+2x^2+2|+C$

✅ Correct Option: 4

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CUET Mathematics 2026 18th May Shift 2 PYQs

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