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Question

For the differential equation in given question find a particular solution satisfying the given condition.​​

x(x21)dydx=1, where y=0 and x=2

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Solution

Given, x(x21)dydx=1
On separating the variables, we get
dy=1x(x21)dxdy=1x(x1)(x+1)dx
On integrating both sides, we get dy=1x(x1)(x+1)dx ....(i)Let 1x(x1)(x+1)=Ax+Bx1+Cx+1 ....(ii) (by partial fraction)
1x(x1)(x+1)=A(x1)(x+1)+Bx(x+1)+Cx(x1)x(x1)(x+1)

1=A(x21)+B(x)(x+1)+C(x)(x1)1=Ax2A+Bx2+Bx+Cx2Cx1=x2(A+B+C)+x(BC)+(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=1B=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(x1)(x+1)=1x+12(x1)+12(x+1)dy=1xdx+121x1dx+121x+1dx
Now from eq. (i) we get
y=log|x|+12log|(x1)|+12log|(x1)|+Cy=12log(x2)+12log|(x1)|+12log|(x+1)|+Cy=12[((x1)(x+1)x2)]+C (log l+log mlog n=log lmn)
y=12[log(x21x2)]+C ...... (vii)
On putting y = 0 and x = 2 in Eq. (vii), we get
0=12[log(22122)]+CC=12log(34)
On substituting the value of C in Eq. (vii), we get
y=12log(x21x2)12log(34)
which is the required particular solution.


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