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Mathematics Question on Differential equations

The integrating factor of the first order differential equation x2(x2−1)dydx+x(x2+1)y=x2−1x^{2}\left(x^{2}-1\right) \frac{dy}{dx}+x\left(x^{2}+1\right)y=x^{2}-1 is

A

exe^x

B

x−1xx-\frac{1}{x}

C

x+1xx+\frac{1}{x}

D

1x2\frac{1}{x^{2}}

Answer

x−1xx-\frac{1}{x}

Explanation

Solution

We have,
x2(x2−1)dydx+x(x2+1)y=x2−1x^{2}\left(x^{2}-1\right) \frac{d y}{d x}+x\left(x^{2}+1\right) y=x^{2}-1
dydx+x2+1x(x2−1)y=1x2\frac{d y}{d x}+\frac{x^{2}+1}{x\left(x^{2}-1\right)} y=\frac{1}{x^{2}}
⇒IF=ex2+1x(x2−1)\Rightarrow I F=e^{\frac{x^{2}+1}{x\left(x^{2}-1\right)}}
∴=e1x2+1x(x−1)(x+1)dx\therefore =e^{1} \frac{x^{2}+1}{x(x-1)(x+1)} d x
Let x2+1x(x−1)(x+1)=Ax+Bx−1+Cx+1\frac{x^{2}+1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1} ⇒x2+1=A(x−1)(x+1)+Bx(x+1)\Rightarrow x^{2}+1=A(x-1)(x+1)+B x(x+1) +Cx(x−1)+C x(x-1)
Put x=0,x=0,
∴1=−A\therefore 1=-A
⇒A=−1\Rightarrow A=-1
Put x=1,x=1,
∴2=2B\therefore 2=2 B
⇒B=1\Rightarrow B=1
Put x=−1,x=-1,
∴2=2C\therefore 2=2 C
⇒C=1\Rightarrow C=1
∴x2+1x(x−1)(x+1)=−1x+1x−1+1x+1\therefore \frac{x^{2}+1}{x(x-1)(x+1)}=\frac{-1}{x}+\frac{1}{x-1}+\frac{1}{x+1}
∴IF=e∫(−1x+1x−1+1x+1)dx\therefore I F=e^{\int\left(\frac{-1}{x}+\frac{1}{x-1}+\frac{1}{x+1}\right) d x}
=e[−log⁡x+log⁡(x−1)+log⁡(x+1)]=e^{[-\log x+\log (x-1)+\log (x+1)]}
=elog(x2−1x)=x2−1x=e^{\text{log}\left(\frac{x^{2}-1}{x}\right)}=\frac{x^{2}-1}{x}
=x−1x=x-\frac{1}{x}