For the differential equation in given question find a particular solution satisfying the given condition.
x(x2−1)dydx=1, where y=0 and x=2
Given, x(x2−1)dydx=1
On separating the variables, we get
dy=1x(x2−1)dx⇒dy=1x(x−1)(x+1)dx
On integrating both sides, we get ∫dy=∫1x(x−1)(x+1)dx ....(i)Let 1x(x−1)(x+1)=Ax+Bx−1+Cx+1 ....(ii) (by partial fraction)
⇒1x(x−1)(x+1)=A(x−1)(x+1)+Bx(x+1)+Cx(x−1)x(x−1)(x+1)
∴1=A(x2−1)+B(x)(x+1)+C(x)(x−1)⇒1=Ax2−A+Bx2+Bx+Cx2−Cx⇒1=x2(A+B+C)+x(B−C)+(−A)
On comparing the coefficients of x2, x and constant terms on both sides, we get
A + B + C = 0 ...... (iii)
B - C = 0 ...... (iv)
and -A = 1 ...... (v)
On putting the value of A in Eq. (iii), we get
B + C = 1 ..... (vi)
On adding Eqs. (iv) and (vi), we get 2B=1⇒B=12
Then, from Eq. (iv), C=12 and B=12
On putting the values of A, B and C in Eq. (ii), we get
1x(x−1)(x+1)=−1x+12(x−1)+12(x+1)⇒∫dy=−∫1xdx+12∫1x−1dx+12∫1x+1dx
Now from eq. (i) we get
y=−log|x|+12log|(x−1)|+12log|(x−1)|+C⇒y=−12log(x2)+12log|(x−1)|+12log|(x+1)|+C⇒y=12[((x−1)(x+1)x2)]+C (∴log l+log m−log n=log lmn)
⇒y=12[log∣∣(x2−1x2)∣∣]+C ...... (vii)
On putting y = 0 and x = 2 in Eq. (vii), we get
∴0=12[log∣∣(22−122)∣∣]+C⇒C=−12log(34)
On substituting the value of C in Eq. (vii), we get
y=12log∣∣(x2−1x2)∣∣−12log(34)
which is the required particular solution.