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:
\(y=\frac{1}{2}log[\frac{4(x-1)(x+1)}{3x^2}]\)
\(y=\frac{1}{2}log[\frac{4(x^2-1)}{3x^2}]\)