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Solve the differential equation:

\(\rm (x^2+1){dy\over dx} = 2xy\)


1. ln y = tan-1 x + c
2. ln y = tan-1 (x2 + 1) + c
3. y = c (x2 + 1)
4. y = cx2 

1 Answer

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Best answer
Correct Answer - Option 3 : y = c (x2 + 1)

Calculation:

Given differential equation 

\(\rm (x^2+1){dy\over dx} = 2xy\)

\(\rm {dy\over y} = {2x\over x^2+1} dx\)

Let x2 = t ⇒ 2x dx = dt

\(\rm {dy\over y} = {dt\over t+1} \)

Integrating both the sides, we get

\(\rm \int {dy\over y} =\int {dt\over t+1} \)

ln y = ln (t + 1) + ln c

ln y = ln [(x2 + 1) c]           [log m + log n = log mn]

y = C (x2 + 1)

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