Find f(x)f(x)

Given that x2∫0x(f(t))3 dt=(∫0xf(t) dt)3x^2\displaystyle\int_{0}^{x} (f(t))^3 \ dt=\left(\displaystyle\int_{0}^{x} f(t) \ dt \right)^3 find all the possible functions f(x)f(x).

#Calculus #FunctionalEquations #Integration #DefiniteIntegral

Note by Pratik Shastri
6 years, 8 months ago

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Comments

We begin with writing :

g(x)=∫0xf(t)dtg(x)=\int _{ 0 }^{ x }{ f(t)dt }

⇒g1(x)=f(x)\Rightarrow {g}_{1}(x)=f(x) where g1(x)=dg(x)dx{g}_{1}(x)=\frac{dg(x)}{dx}

Our equation becomes x2∫0xf(t)3dt=g(x)3{x}^{2}\int _{ 0 }^{ x }{ { f(t) }^{ 3 }dt } = { g(x) }^{ 3 } (i)

Differentiating both sides with respect to x we get :

2x∫0xf(t)3dt+x2(g1(x))3=3(g(x)2)(g′(x))2x\int _{ 0 }^{ x }{ { f(t) }^{ 3 }dt }+{x}^{2} {({g}_{1}(x))}^{3}=3({g(x)}^{2})({g}^{'}(x)) (ii)

Using (i) we write (ii) as :

2(g(x)3x)+x2(g1(x))3=3(g(x)2)(g1(x))2(\frac { { g(x })^{ 3 } }{ x }) + {x}^{2}{({g}_{1}(x))}^{3}=3({g(x)}^{2})({g}_{1}(x)) (iii)

After some rearranging we get :

(xg1(x))3−3g(x)2(xg1(x))+2g(x)3=0{(x{g}_{1}(x))}^{3}-3{g(x)}^{2}(x{g}_{1}(x))+2{g(x)}^{3}=0

Dividing both sides by g(x)3{g(x)}^{3} we get :

(xg1(x)g(x))3−3(xg1(x)g(x))+2=0{(\frac { x{ g }_{ 1 }(x) }{ g(x) })}^{3}-3(\frac { x{ g }_{ 1 }(x) }{ g(x) }) +2 =0

Take xg1(x)g(x)=y\frac { x{ g }_{ 1 }(x) }{ g(x) } =y to get our equation as :

y3−3y+2=0{y}^{3}-3y+2=0

Factorising this we get :

(y+2)(y−1)2=0(y+2){(y-1)}^{2}=0

⇒y=−2,1\Rightarrow y=-2,1

We will deal both these cases seperately :

Case-1,y=1y=1

xg1(x)g(x)=1\frac { x{ g }_{ 1 }(x) }{ g(x) } =1

⇒∫dg(x)g(x)=∫dxx\Rightarrow \int { \frac { dg(x) }{ g(x) } } =\int { \frac { dx }{ x } }

⇒g(x)=Cx\Rightarrow g(x)=Cx,

Differentiating both sides we get :

f(x)=Cf(x)=C (iv)

Case-2,y=−2y=-2

xg1(x)g(x)=−2\frac { x{ g }_{ 1 }(x) }{ g(x) } =-2

⇒∫dg(x)g(x)=∫−2dxx\Rightarrow \int { \frac { dg(x) }{ g(x) } } =\int { \frac {-2dx }{ x } }

⇒g(x)=Cx2\Rightarrow g(x)=\frac{C}{{x}^{2}}

But since g(0)=0g(0)=0 hence no value of CC exists.

So the only function that exists is f(x)=Cf(x)=C

Ronak Agarwal - 6 years, 8 months ago
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