Question

For the differential equations, find a particular solution satisfying the given condition:$x(x^2-1)\frac{dy}{dx}=1;y=0$ when x=2

Answer 1

$x(x^2-1)\frac{dy}{dx}=1$
$⇒dy=\frac{dx}{x(x^2-1)}$
$⇒dy=\frac{1}{x(x-1)(x+1)}dx$
Integrating both sides,we get:
$∫dy=∫\frac{1}{x(x-1)(x+1)}dx...(1)$
Let $\frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}. ...(2)$
$⇒\frac{1}{x(x-1)(x+1)}=A(x-1)(x+1)+Bx(x+1)+\frac{Cx(x-1)}{x(x-1)(x+1)}$
$=(A+B+C)x^2+(B-C)x-\frac{A}{x(x-1)(x+1)}$
Comparing the coefficients of x2,x,and constant,we get:
$A=-1$
$B-C=0$
$A+B+C=0$
Solving these equations,we get $B=\frac{1}{2}$and $C=\frac{1}{2}$.
Substituting the values of A,B,and C in equation(2),we get:
$\frac{12}{x(x-1)(x+1)}=\frac{-1}{x}+\frac{1}{2}(x-1)+\frac{1}{2}(x+1)$
Therefore,equation(1)becomes:
$∫dy=-∫\frac{1}{x}dx+\frac{1}{2}∫\frac{1}{x}-1dx+\frac{1}{2}∫\frac{1}{x}+1dx$
$⇒y=-logx+\frac{1}{2}log(x-1)+\frac{1}{2}log(x+1)+logk$
$⇒y=\frac{1}{2}log[\frac{k^2(x-1)(x+1)}{x^2}]...(3)$
Now,y=0 when x=2.
$⇒0=\frac{1}{2}log[\frac{k^2(2-1)(2+1)}{4}]$
$⇒log(\frac{3k^2}{4})=0$
$⇒\frac{3k^2}{4}=1$
$⇒3k^2=4$
$⇒k^2=\frac{4}{3}$
Substituting the value of $k^2$ in eqation(3),we get:
$y=\frac{1}{2}log[\frac{4(x-1)(x+1)}{3x^2}]$
$y=\frac{1}{2}log[\frac{4(x^2-1)}{3x^2}]$