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Let y=f(x) be a real-valued differentiable function on the set of all real numbers R such that f(1)=1. If f(x) satisfies xf(x)=x2+f(x)2, then the area enclosed by y=f(x) with x-axis between ordinates x=0 and x=3 is 


Solution

xf(x)=x2+f(x)2
y=f(x), so dydx=f(x)
Now, dydx+(1x)y=x2x 
which is a linear differential equation.
I.F.=exp(1xdx)=elnx=1x

The general solution is
y(1x)=(x2x)1xdx
yx=x+2x+C
As y(1)=1C=2
yx=x+2x2
f(x)=x22x+2=(x1)2+1


Required area
=30((x1)2+1)dx
=[13(x1)3]30+[x]30
=(83+13)+3
=3+3=6

 

Mathematics

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