Question

Find the particular solution of differential equal $x\left(x^2-1\right)\frac{dy}{dx}=1;y=0$ and $x=2.$

Answer 1

$x(x^2-1)dy=dx$

$dy=\frac{dx}{x(x^2-1)}$

Integrating on both sides

$\int dy=\int \frac{dx}{x(x^2-1)}$

$y=\int \frac{dx}{x(x+1)(x-1)}$ …………..$\left(1\right)$

We can write

$\frac{1}{x(x+1)(x-1)}=\frac{A}{x}+\frac{B}{(x+1)}+\frac{C}{(x-1)}$

$

By cancelling the denominators

$1=A(x-1)(x+1)+Bx(x-1)+C(x+1).x$

Putting $x=0$

$1=A(0-1)(0+1)+B.0(0-1)+C.0(0+1)$

$1=-A$

$A=-1$

Similarly putting $x=-1$

$1=A(-1-1)(-1+1)+B(-1)(-1-1)+C(-1)(-1+1)$

$1=A(-2)\left(0\right)+B(-1)(-2)+C:0$

$1=0+2B+0$

$2B=1$

$B1/2$

Putting $x=1$

$1=A(1-1)(1+1)+B\left(1\right)(1-1)+C.1(1+1)$

$1=20.A+0.B+2C$

$1=2C$

$C=1/2$

Therefore $\frac{1}{x(x+1)(x-1)}=\frac{-1}{x}+\frac{1/2}{(x+1)}-\frac{1/2}{(x-1)}$

Now, from $\left(1\right)$

$y=\int \frac{1}{\left(x\right(x+1\left)\right(x-1)}\cdot dx$

$=\int \frac{1}{x}+dx+1/2\int \frac{dx}{x+1}+\frac{1}{2}\int \frac{dx}{x-1}$

$=log|x|+\frac{1}{2}.log|x+1|+\frac{1}{2}log|x-1|+c$

$=\frac{-2}{2}log|x|+\frac{1}{2}log|x+1|+\frac{1}{2}log|x-1|+c$

$=\frac{1}{2}[-2log|x|-2+log|(x+1\left)\right(x-1\left)|\right]+c$

$=\frac{1}{2}\left[log{x}^{-2}log|\right(x+1\left)\right(x-1\left)|\right]+c$

$=\frac{1}{2}\left[log|x^{-2}\right(x^2-1\left)|\right]+c$ $[\because loga+logb-logab]$

$=\frac{1}{2}log\left|\frac{x^2-1}{x^2}\right|+c$

$\therefore y=\frac{1}{2}log\left|\frac{x^2-1}{x^2}\right|+c$ …………..$\left(1\right)$

Given that

$x=2;y=0$

Substituting values in $\left(1\right)$, we get

$0=\frac{1}{2}log\left|\frac{2^2-1}{2^2}\right|+c$

$0=\frac{1}{2}log\frac{3}{4}+c$

$c=-\frac{1}{2}log\frac{3}{4}$

Putting value of c in $\left(1\right)$

$y=\frac{1}{2}log\left|\frac{x^2-1}{x^2}\right|-\frac{1}{2}log3/4$.