NEED HELP WITH EXERCISE

NEED HELP WITH EXERCISE

Postby Guest » Fri Jul 24, 2020 8:03 pm

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

Prove that: $f(x)=x+\sqrt{x^2+1}$
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Re: NEED HELP WITH EXERCISE

Postby HallsofIvy » Sat Jul 25, 2020 10:36 pm

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

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

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

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

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

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

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Re: NEED HELP WITH EXERCISE

Postby Guest » Sun Jul 26, 2020 11:02 am

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

Is it correct if I write it this way?

[tex]\frac{f(x)}{f(x)-x}[/tex]]=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
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