Mixing
if, $y=lnleft(frac{x}{1-x}right)$ prove $x^3frac{d^2y}{dx^2}= left(xfrac{dy}{dx}-yright)^2$
$begingroup$
If $y=lnleft(frac{x}{1-x}right)$, prove
$x^3frac{d^2y}{dx^2}= left(xfrac{dy}{dx}-yright)^2$
My Try
I was able to differentiate this expression twice without a truoble, but I'm having difficulty in obtaining 'y' in the second derivative expression. Please help.
derivatives
edited Jan 14 at 5:03
coreyman317
772420
asked Jan 14 at 3:56
emilemil
431410
$endgroup$
add a comment |
$begingroup$
If $y=lnleft(frac{x}{1-x}right)$, prove
$x^3frac{d^2y}{dx^2}= left(xfrac{dy}{dx}-yright)^2$
My Try
I was able to differentiate this expression twice without a truoble, but I'm having difficulty in obtaining 'y' in the second derivative expression. Please help.
derivatives
edited Jan 14 at 5:03
coreyman317
772420
asked Jan 14 at 3:56
emilemil
431410
$endgroup$
add a comment |
$begingroup$
If $y=lnleft(frac{x}{1-x}right)$, prove
$x^3frac{d^2y}{dx^2}= left(xfrac{dy}{dx}-yright)^2$
My Try
I was able to differentiate this expression twice without a truoble, but I'm having difficulty in obtaining 'y' in the second derivative expression. Please help.
derivatives
edited Jan 14 at 5:03
coreyman317
772420
asked Jan 14 at 3:56
emilemil
431410
$endgroup$
If $y=lnleft(frac{x}{1-x}right)$, prove
$x^3frac{d^2y}{dx^2}= left(xfrac{dy}{dx}-yright)^2$
My Try
I was able to differentiate this expression twice without a truoble, but I'm having difficulty in obtaining 'y' in the second derivative expression. Please help.
derivatives
derivatives
edited Jan 14 at 5:03
coreyman317
772420
asked Jan 14 at 3:56
emilemil
431410
edited Jan 14 at 5:03
coreyman317
772420
asked Jan 14 at 3:56
emilemil
431410
edited Jan 14 at 5:03
coreyman317
772420
edited Jan 14 at 5:03
coreyman317
772420
edited Jan 14 at 5:03
coreyman317
772420
772420
asked Jan 14 at 3:56
emilemil
431410
asked Jan 14 at 3:56
emilemil
431410
asked Jan 14 at 3:56
emilemil
431410
431410
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is an exercise showing your given function satisfies the given differential equation!
According to your equation, we need $frac{d^2y}{dx^2}$ and $frac{dy}{dx}$: $$frac{dy}{dx}=left(lnleft(frac{x}{1-x}right)right)^{'}=frac{1}{frac{x}{1-x}}cdotleft(frac{x}{1-x}right)^{'
}=frac{1-x}{x}cdotfrac{(1-x)-x(-1)}{(1-x)^2}=frac{1}{x(1-x)}=frac{1}{x}-frac{1}{x-1}$$ $$frac{d^2y}{dx^2}=left(frac{1}{x}-frac{1}{x-1}right)^{'}=-frac{1}{x^2}+frac{1}{(x-1)^2}$$
Plugging all of this into $x^3cdotfrac{d^2y}{dx^2}=left(xfrac{dy}{dx}-yright)^2$: $$x^3left(-frac{1}{x^2}+frac{1}{(x-1)^2}right)=^?left(xleft(frac{1}{x}-frac{1}{x-1}right)-lnleft(frac{x}{1-x}right)right)^2$$ $$-x+frac{x^3}{(x-1)^2}=^?left(1-frac{x}{x-1}-lnleft(frac{x}{1-x}right)right)^2$$
However the above relation can't be true since a rational function and polynomial are on the left while a logarithm is on the RHS.
answered Jan 14 at 4:28
coreyman317coreyman317
772420
$endgroup$
add a comment |
$begingroup$
I do not think the statement to be proven is true. We clearly have
$$y=ln x-ln(1-x),$$
and so
$$y'=frac 1x+frac{1}{1-x},,,,,y''=-frac{1}{x^2}+frac{1}{(1-x)^2}.$$
Let $x=1/2$. Then it is easy to see that the LHS is
$$x^3 y''=0$$
but the RHS is
$$(xy'-y)^2=left(frac 12times 4-0right)^2=4.$$
answered Jan 14 at 4:18
user146512user146512
636
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
This is an exercise showing your given function satisfies the given differential equation!
According to your equation, we need $frac{d^2y}{dx^2}$ and $frac{dy}{dx}$: $$frac{dy}{dx}=left(lnleft(frac{x}{1-x}right)right)^{'}=frac{1}{frac{x}{1-x}}cdotleft(frac{x}{1-x}right)^{'
}=frac{1-x}{x}cdotfrac{(1-x)-x(-1)}{(1-x)^2}=frac{1}{x(1-x)}=frac{1}{x}-frac{1}{x-1}$$ $$frac{d^2y}{dx^2}=left(frac{1}{x}-frac{1}{x-1}right)^{'}=-frac{1}{x^2}+frac{1}{(x-1)^2}$$
Plugging all of this into $x^3cdotfrac{d^2y}{dx^2}=left(xfrac{dy}{dx}-yright)^2$: $$x^3left(-frac{1}{x^2}+frac{1}{(x-1)^2}right)=^?left(xleft(frac{1}{x}-frac{1}{x-1}right)-lnleft(frac{x}{1-x}right)right)^2$$ $$-x+frac{x^3}{(x-1)^2}=^?left(1-frac{x}{x-1}-lnleft(frac{x}{1-x}right)right)^2$$
However the above relation can't be true since a rational function and polynomial are on the left while a logarithm is on the RHS.
answered Jan 14 at 4:28
coreyman317coreyman317
772420
$endgroup$
add a comment |
$begingroup$
This is an exercise showing your given function satisfies the given differential equation!
According to your equation, we need $frac{d^2y}{dx^2}$ and $frac{dy}{dx}$: $$frac{dy}{dx}=left(lnleft(frac{x}{1-x}right)right)^{'}=frac{1}{frac{x}{1-x}}cdotleft(frac{x}{1-x}right)^{'
}=frac{1-x}{x}cdotfrac{(1-x)-x(-1)}{(1-x)^2}=frac{1}{x(1-x)}=frac{1}{x}-frac{1}{x-1}$$ $$frac{d^2y}{dx^2}=left(frac{1}{x}-frac{1}{x-1}right)^{'}=-frac{1}{x^2}+frac{1}{(x-1)^2}$$
Plugging all of this into $x^3cdotfrac{d^2y}{dx^2}=left(xfrac{dy}{dx}-yright)^2$: $$x^3left(-frac{1}{x^2}+frac{1}{(x-1)^2}right)=^?left(xleft(frac{1}{x}-frac{1}{x-1}right)-lnleft(frac{x}{1-x}right)right)^2$$ $$-x+frac{x^3}{(x-1)^2}=^?left(1-frac{x}{x-1}-lnleft(frac{x}{1-x}right)right)^2$$
However the above relation can't be true since a rational function and polynomial are on the left while a logarithm is on the RHS.
answered Jan 14 at 4:28
coreyman317coreyman317
772420
$endgroup$
add a comment |
$begingroup$
This is an exercise showing your given function satisfies the given differential equation!
According to your equation, we need $frac{d^2y}{dx^2}$ and $frac{dy}{dx}$: $$frac{dy}{dx}=left(lnleft(frac{x}{1-x}right)right)^{'}=frac{1}{frac{x}{1-x}}cdotleft(frac{x}{1-x}right)^{'
}=frac{1-x}{x}cdotfrac{(1-x)-x(-1)}{(1-x)^2}=frac{1}{x(1-x)}=frac{1}{x}-frac{1}{x-1}$$ $$frac{d^2y}{dx^2}=left(frac{1}{x}-frac{1}{x-1}right)^{'}=-frac{1}{x^2}+frac{1}{(x-1)^2}$$
Plugging all of this into $x^3cdotfrac{d^2y}{dx^2}=left(xfrac{dy}{dx}-yright)^2$: $$x^3left(-frac{1}{x^2}+frac{1}{(x-1)^2}right)=^?left(xleft(frac{1}{x}-frac{1}{x-1}right)-lnleft(frac{x}{1-x}right)right)^2$$ $$-x+frac{x^3}{(x-1)^2}=^?left(1-frac{x}{x-1}-lnleft(frac{x}{1-x}right)right)^2$$
However the above relation can't be true since a rational function and polynomial are on the left while a logarithm is on the RHS.
answered Jan 14 at 4:28
coreyman317coreyman317
772420
$endgroup$
This is an exercise showing your given function satisfies the given differential equation!
According to your equation, we need $frac{d^2y}{dx^2}$ and $frac{dy}{dx}$: $$frac{dy}{dx}=left(lnleft(frac{x}{1-x}right)right)^{'}=frac{1}{frac{x}{1-x}}cdotleft(frac{x}{1-x}right)^{'
}=frac{1-x}{x}cdotfrac{(1-x)-x(-1)}{(1-x)^2}=frac{1}{x(1-x)}=frac{1}{x}-frac{1}{x-1}$$ $$frac{d^2y}{dx^2}=left(frac{1}{x}-frac{1}{x-1}right)^{'}=-frac{1}{x^2}+frac{1}{(x-1)^2}$$
Plugging all of this into $x^3cdotfrac{d^2y}{dx^2}=left(xfrac{dy}{dx}-yright)^2$: $$x^3left(-frac{1}{x^2}+frac{1}{(x-1)^2}right)=^?left(xleft(frac{1}{x}-frac{1}{x-1}right)-lnleft(frac{x}{1-x}right)right)^2$$ $$-x+frac{x^3}{(x-1)^2}=^?left(1-frac{x}{x-1}-lnleft(frac{x}{1-x}right)right)^2$$
However the above relation can't be true since a rational function and polynomial are on the left while a logarithm is on the RHS.
answered Jan 14 at 4:28
coreyman317coreyman317
772420
answered Jan 14 at 4:28
coreyman317coreyman317
772420
answered Jan 14 at 4:28
coreyman317coreyman317
772420
answered Jan 14 at 4:28
coreyman317coreyman317
772420
772420
add a comment |
add a comment |
$begingroup$
I do not think the statement to be proven is true. We clearly have
$$y=ln x-ln(1-x),$$
and so
$$y'=frac 1x+frac{1}{1-x},,,,,y''=-frac{1}{x^2}+frac{1}{(1-x)^2}.$$
Let $x=1/2$. Then it is easy to see that the LHS is
$$x^3 y''=0$$
but the RHS is
$$(xy'-y)^2=left(frac 12times 4-0right)^2=4.$$
answered Jan 14 at 4:18
user146512user146512
636
$endgroup$
add a comment |
$begingroup$
I do not think the statement to be proven is true. We clearly have
$$y=ln x-ln(1-x),$$
and so
$$y'=frac 1x+frac{1}{1-x},,,,,y''=-frac{1}{x^2}+frac{1}{(1-x)^2}.$$
Let $x=1/2$. Then it is easy to see that the LHS is
$$x^3 y''=0$$
but the RHS is
$$(xy'-y)^2=left(frac 12times 4-0right)^2=4.$$
answered Jan 14 at 4:18
user146512user146512
636
$endgroup$
add a comment |
$begingroup$
I do not think the statement to be proven is true. We clearly have
$$y=ln x-ln(1-x),$$
and so
$$y'=frac 1x+frac{1}{1-x},,,,,y''=-frac{1}{x^2}+frac{1}{(1-x)^2}.$$
Let $x=1/2$. Then it is easy to see that the LHS is
$$x^3 y''=0$$
but the RHS is
$$(xy'-y)^2=left(frac 12times 4-0right)^2=4.$$
answered Jan 14 at 4:18
user146512user146512
636
$endgroup$
I do not think the statement to be proven is true. We clearly have
$$y=ln x-ln(1-x),$$
and so
$$y'=frac 1x+frac{1}{1-x},,,,,y''=-frac{1}{x^2}+frac{1}{(1-x)^2}.$$
Let $x=1/2$. Then it is easy to see that the LHS is
$$x^3 y''=0$$
but the RHS is
$$(xy'-y)^2=left(frac 12times 4-0right)^2=4.$$
answered Jan 14 at 4:18
user146512user146512
636
answered Jan 14 at 4:18
user146512user146512
636
answered Jan 14 at 4:18
user146512user146512
636
answered Jan 14 at 4:18
user146512user146512
636
636
add a comment |
add a comment |
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