Let y=y(x) be the solution of the differential equation, `(x^(2)+1)^(2)dy/dx+2x(x^(2)+1)y=1` such that y(0) =0. If `sqrta y(1)=pi/32`,then tha value of 'a' is (a) 1/4 (b) 1/2 (c) 1 (d) 1/16

Options

A. (a) 1/4
B. (b) 1/2
C. (c) 1
D. (d) 1/16

Answer

(d)

Explanation

Given differential equation is
`(x^(2)+1)^(2) dy/dx+2x(x^(2)+1)y=1`
` rArr dy/dx=((2x)/(1+x^(2)))y=1/((1=x^(2))^(2)`
[dividing each term by `(1+x^(2))^(2)`] … (i)
this is a linear differential equation of the form
`dy/dx+P cdot y= Q`
Here, `P=(2x)/((1+x^(2) ) ) and Q=1/(1+x^(2))^(2)`
`therefore Integrating Factor (IF)= e^(int(2x)/(1+x^(2))dx`
`=e^(ln(1+x^(2)))=(1+x^(2))`
and required solution of differential Eq. (i) is given by
`y cdot (IF)= int Q(IF)dx=C`
` rArr y(1+x^(2))= int 1/((1+x^(2))^(2)) (1+x^(2)) dx+C`
`rArr y(1+x^(2))=intdx/(1+x^(2))+C`
`rArr y(1+x^(2))=tan^(-1)(x)+C`
`therefore y(0)=0`
`therefore C=0`
`therefore y(1+x^(2))= tan ^(-1) x [therefore C=0]`
`rArr y=(tan^(-1)x)/(1+x^(2))`
`rArr y=(tan^(_1)x)/(1+x^(2))`
`rArr sqrtay =sqrta((tan^(-1)x)/(1+x^(2)))`
[multiplying both sides by `sqrt(a)`]
Now, `at x = 1`
` sqrta y (1)=sqrta((tan^(-1)(1))/(1+1))=sqrta (pi/4)/2=(sqrtapi)/8=pi/32` (given)
`therefore sqrta= 1/4 rArr a =1/16`