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Question

Suppose y=f(x) is the solution of the differential equation (1x2)dydx=(1x2)(x4+2x)+xy for x(a,b), where f(0)=0. Then

A
f(x) is an even function.
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B
f(x) is an odd function.
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C
ba=2
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D
1/21/2f(x) dx=π634
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Solution

The correct options are
C ba=2
D 1/21/2f(x) dx=π634
Given : (1x2)dydx=(1x2)(x4+2x)+xy
dydx+xyx21=x4+2x1x2
For the square root to be defined,
1x2>0x(1,1)

Comparing with dydx+P(x)y=Q(x)
I.F.=eP(x)dx
P(x)dx=xx21dx
Assuming x21=t2x dx=dt, then
P(x)dx=12dttP(x)dx=12ln|x21|
As x(1,1), so
P(x)dx=12ln(1x2)I.F.=1x2

Solution is :
y1x2=(x4+2x) dxy1x2=x55+x2+Cf(0)=0, C=0y=f(x)=x55+x21x2
Clearly, f(x) is neither even nor odd function.
f(x) is defined x(1,1)
Therefore, ba=2

I=1/21/2x55+x21x2 dxI=1/21/2x21x2dx (x5 is odd function)I=21/20x21x2dx

Taking x=sinθdx=cosθ dθ
I=2π/60sin2θ dθI=π/60(1cos2θ) dθI=[θsin2θ2]π/60I=π634

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