IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

ramyadiamond

Since, f(x) is dfferentiable, hence, use L'Hosptials Rule in the limit.

_{[ t]}_{[x ]} t².f(x)-x².f(t)/t-x =1

Differentiating w.r.t t, we get

_{[t ]}_{[ x]} 2t. f(x) -x².f^'(t) =1

Now apply the limit, and u'll get

2x.f(x) -x².f^'(x) =1

Now i'm writing f(x)=y, it'll be easier to mention.

Now,

2xy -x².dy/dx =1

x²dy/dx -2xy =-1

dy/dx -2y/x = -1/x²

U can compare this with the form,

dy/dx +Py =Q

where integrating factor (I.F) can be found by

I.F= e^ P dx

Hence, for the above differential equation,

I.F = e^ (-2/x) dx

= 1/x²

We know,

y(I.F.) = Q. (I.F.) dx

y(1/x²) = (-1/x²)*(1/x²) dx

Hence,

y (1/x²) = 1/3x³ +c

where c is the const. of integration.

y = 1/3x +cx²

Given that, for x=1, y=1

Hence,

1= 1/3(1) +c(1)

Therefore, c=2/3

Hence the function,

y=f(x) = 1/3x+2x²/3

Answer is (a)

Hope this helps!

Cheerrsss!!!