Question

# 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=x−2x  which is a linear differential equation. I.F.=exp(∫−1xdx)=e−lnx=1x The general solution is y(1x)=∫(x−2x)1xdx ∴yx=x+2x+C As y(1)=1⇒C=−2 ∴yx=x+2x−2 ⇒f(x)=x2−2x+2=(x−1)2+1 ∴ Required area =∫30((x−1)2+1)dx =[13(x−1)3]30+[x]30 =(83+13)+3 =3+3=6  Mathematics

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