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Question: Find the integral factor of equation \((x^{2} + 1)\frac{dy}{dx} + 2xy = x^{2} - 1\)...

Find the integral factor of equation (x2+1)dydx+2xy=x2−1(x^{2} + 1)\frac{dy}{dx} + 2xy = x^{2} - 1

A

x2+1x^{2} + 1

B

2xx2+1\frac{2x}{x^{2} + 1}

C

x2−1x2+1\frac{x^{2} - 1}{x^{2} + 1}

D

None of these

Answer

x2+1x^{2} + 1

Explanation

Solution

Given equation may be written as dydx+2xx2+1y=x2−1x2+1\frac{dy}{dx} + \frac{2x}{x^{2} + 1}y = \frac{x^{2} - 1}{x^{2} + 1}

Comparing with dydx+Py=Q\frac{dy}{dx} + Py = Q,

P=2xx2+1P = \frac{2x}{x^{2} + 1}

I.F.=e∫Pdx=e∫2xdx1+x2=eln⁡(1+x2)=1+x2= e^{\int_{}^{}{Pdx}} = e^{\int_{}^{}\frac{2xdx}{1 + x^{2}}} = e^{\ln(1 + x^{2})} = 1 + x^{2}