NEED HELP WITH EXERCISE

NEED HELP WITH EXERCISE

f:R$$\rightarrow$$R, f(0)=1 , x/f(x)+1/f'(x)=1 for every $x\in R$:

Prove that: $f(x)=x+\sqrt{x^2+1}$
Guest

Re: NEED HELP WITH EXERCISE

The equation is $$\frac{x}{f(x)}+ \frac{1}{f'(x)}= 1$$ with the initial condition f(0)= 1.
I am going to use "y" rather than "f" to make it a little easier to type: $$\frac{x}{y}+ \frac{1}{y'}= 1$$.

Multiply both sides by yy' to get rid of the fractions. $$xy'+ y= yy'$$. Subtract xy' from both sides:
$$y= yy'- xy'= (y- x)y'$$. Divide both sides by y- x: $$y'= \frac{y}{y- x}$$.

Let u= y- x. Then y= u+ x and y'= u'+ 1 so the equation becomed $$u'+ 1= \frac{u+ x}{u}= 1+ \frac{x}{u}$$.

So $$u'= \frac{du}{dx}= \frac{x}{u}$$. Separate variables- $$udu= xdx$$ and integrate- $$\frac{u^2}{2}= \frac{x^2}{2}+ C$$ or $$u^2= x^2+ 2C$$.

Since u= y- x, $$(y- x)^2= y^2- 2xy+x^2= x^2+ 2C$$.
The two $$x^2$$ terms cancel so $$y^2- 2xy= 2C$$. y(0)= 1 so $$1^2- 2(0)(1)= 1= 2C$$. Then $$y^2- 2xy- 1= 0$$.

Solve that for y by "completing the square"- $$y^2- 2xy+ x^2- x^2- 1= 0$$. $$y^2- 2xy+ x^2= (y- x)^2= x^2+ 1$$. Take the square root: $$y- x= \pm\sqrt{x^2+1}$$ so $$y= x\pm \sqrt{x^2+ 1}$$
$$y= x- \sqrt{x^2- 1}$$ however, does not satisfy the condition that y(0)= 1 so y= f(x) must equal $$x+ \sqrt{x^2+ 1}$$.

HallsofIvy

Posts: 259
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 96

Re: NEED HELP WITH EXERCISE

Damn, I feel so dumb now, it didn't occur to me to set a new equation

Is it correct if I write it this way?

$$\frac{f(x)}{f(x)-x}$$]=f'(x), set: g(x)=f(x)-x (1) , g'(x)=f'(x)-1, f(x)=g(x)+x

So the equation is rewritten as:

\frac{g(x)+x}{g(x)}=g'(x)+1 \Leftrightarrow

\frac{g(x)+x}{g(x)}=g'(x)+\frac{g(x)}{g(x)} \Leftrightarrow

\frac{g(x)+x}{g(x)}-\frac{g(x)}{g(x)}=g'(x)\Leftrightarrow

\frac{g(x)+x-g(x)}{g(x)}=g'(x) \Leftrightarrow

\frac{x}{g(x)}=g'(x)

Multiply both sides by g(x)

g(x)g'(x)=x

Divide both sides by 2

\frac{g(x)g'(x)}{2}=\frac{x}{2}\Rightarrow

(\frac{(g(x)^{2})}{2})'=(\frac{x^2}{2})'

The integrated equation is:

\frac{g^2(x)}{2}= (\frac{x^2}{2})+c (2)

For x=0 we have:

\frac{g^2(0)}{2}=\frac{0}{2}+c \Rightarrow (1)

\frac{f^2(0)-2f(0)\cdot0+0}{2}=c

\frac{f^2(0)}{2}=c , f(0)=1

c=\frac{1}{2}

So equation (2) becomes:

\frac{g^2(x)}{2}= (\frac{x^2}{2})+ \frac{1}{2} or g^2(x)=x^2+1

We take the square root on both sides

\sqrt{g^2(x)}=\sqrt{x^2+1}

|g(x)|= \sqrt{x^2+1}

Replace g(x) with f(x)-x

|f(x)-x|= \sqrt{x^2+1}

If f(x)-x>0, f(x)= \sqrt{x^2+1}+x

If f(x)-x<0, f(x)= x-\sqrt{x^2+1} FALSE since f(0)=-1 in this case, this does not satisfy our initial condition of f(0)=1
Guest