Mixing
Differential Equations - Is the solution to $(e^y+1)^2e^{-y}dx+(e^x+1)^6e^{-x}dy=0$, $y=ln(-5(e^x+1)^5+c)$?
$begingroup$
This is the problem. It is question 4. I typed the answer in explicit form, but my professor had said it is incorrect. Could someone check over my work, because I think I did have it correct. My Work
Because someone asked for my work due to bad handwriting:
$$(e^y+1)^2e^{-y}dx+(e^x+1)^6e^{-x}dy=0$$
Trying to seperate variables:
$$(e^y+1)^2(e^{-y})dx=-(e^x+1)^6(e^{-x})dy$$
$$frac{-e^x}{(e^x+1)^6}dx=frac{e^y}{(e^y+1)^2}dy$$
$$int{frac{-e^x}{(e^x+1)^6}dx}=int{frac{e^y}{(e^y+1)^2}dy}$$
Flip:
$$int{frac{e^y}{(e^y+1)^2}dy}=int{frac{-e^x}{(e^x+1)^6}dx}$$
where $u = (e^y+1), du =e^ydy$ and $v = (e^x+1), dv=e^xdx$
$$int{u^{-2}dy}=-int{v^{-6}}dx$$
$$-u^{-1} + c_1=frac{1}{5}v^{-5}+c_2$$
Now, substitute:
$$-frac{1}{e^y+1}+c_1=frac{1}{5(e^x+1)^5}+c_2$$
The $c$ gets absorbed (also, this is the implicit solution)
$$-frac{1}{e^y+1}=frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
But $-c_3$ is the same as $+c_3$
$$-5(e^x+1)^5+c_3=e^y+1$$
$$-5(e^x+1)^5+c_3=e^y$$
Take the natural log of both sides:
$$ln(-5(e^x+1)^5+c_3)=ln(e^y)$$
$$lnleft(-5(e^x+1)^5+c_3right)=y$$
This is the answer I got. It took long to convert this to Latex, so there might be some discontinuities in the work, but the last answer is for sure what I got.
ordinary-differential-equations
edited Jan 20 at 22:27
El borito
666216
asked Jan 20 at 21:15
Kenneth DangKenneth Dang
136
$endgroup$
|
show 1 more comment
$begingroup$
This is the problem. It is question 4. I typed the answer in explicit form, but my professor had said it is incorrect. Could someone check over my work, because I think I did have it correct. My Work
Because someone asked for my work due to bad handwriting:
$$(e^y+1)^2e^{-y}dx+(e^x+1)^6e^{-x}dy=0$$
Trying to seperate variables:
$$(e^y+1)^2(e^{-y})dx=-(e^x+1)^6(e^{-x})dy$$
$$frac{-e^x}{(e^x+1)^6}dx=frac{e^y}{(e^y+1)^2}dy$$
$$int{frac{-e^x}{(e^x+1)^6}dx}=int{frac{e^y}{(e^y+1)^2}dy}$$
Flip:
$$int{frac{e^y}{(e^y+1)^2}dy}=int{frac{-e^x}{(e^x+1)^6}dx}$$
where $u = (e^y+1), du =e^ydy$ and $v = (e^x+1), dv=e^xdx$
$$int{u^{-2}dy}=-int{v^{-6}}dx$$
$$-u^{-1} + c_1=frac{1}{5}v^{-5}+c_2$$
Now, substitute:
$$-frac{1}{e^y+1}+c_1=frac{1}{5(e^x+1)^5}+c_2$$
The $c$ gets absorbed (also, this is the implicit solution)
$$-frac{1}{e^y+1}=frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
But $-c_3$ is the same as $+c_3$
$$-5(e^x+1)^5+c_3=e^y+1$$
$$-5(e^x+1)^5+c_3=e^y$$
Take the natural log of both sides:
$$ln(-5(e^x+1)^5+c_3)=ln(e^y)$$
$$lnleft(-5(e^x+1)^5+c_3right)=y$$
This is the answer I got. It took long to convert this to Latex, so there might be some discontinuities in the work, but the last answer is for sure what I got.
ordinary-differential-equations
edited Jan 20 at 22:27
El borito
666216
asked Jan 20 at 21:15
Kenneth DangKenneth Dang
136
$endgroup$
$begingroup$
You can't be serious!!! Your "work" is impossible to read. How about writing one equation per line? Or better still, how about transcribing your work in TeX?
$endgroup$
– Bernard Massé
Jan 20 at 21:23
$begingroup$
If you think your answer is the right one, just derive and plug it in the equation. If you come up with an equality, then your answer is good. Or use Wolfram Alpha to verify your answer.
$endgroup$
– Bernard Massé
Jan 20 at 21:30
1
$begingroup$
No, $-frac1a=frac1b+c$ is not equivalent to $-b-c=a$. Not even with a different constant.
$endgroup$
– LutzL
Jan 20 at 22:22
1
$begingroup$
Why? Wouldn't that just be $-frac{b}{a}=1-bc$ $implies$ $-frac{b}{a}=1-c$ $implies$ $-frac{b}{a}+c=1$ $implies$ $-b+ac=a$ $implies$ $-b+c=a$? or as you said $-b-c=a$?
$endgroup$
– Kenneth Dang
Jan 20 at 22:29
1
$begingroup$
$a$ and $b$ are the variable terms containing $x$ and $y$, $c$ is the integration constant. Multiplying by $ab$ gives $-b=a+cab$, and the last term is not constant, can thus not be replaced with a new constant.
$endgroup$
– LutzL
Jan 20 at 23:13
|
show 1 more comment
$begingroup$
This is the problem. It is question 4. I typed the answer in explicit form, but my professor had said it is incorrect. Could someone check over my work, because I think I did have it correct. My Work
Because someone asked for my work due to bad handwriting:
$$(e^y+1)^2e^{-y}dx+(e^x+1)^6e^{-x}dy=0$$
Trying to seperate variables:
$$(e^y+1)^2(e^{-y})dx=-(e^x+1)^6(e^{-x})dy$$
$$frac{-e^x}{(e^x+1)^6}dx=frac{e^y}{(e^y+1)^2}dy$$
$$int{frac{-e^x}{(e^x+1)^6}dx}=int{frac{e^y}{(e^y+1)^2}dy}$$
Flip:
$$int{frac{e^y}{(e^y+1)^2}dy}=int{frac{-e^x}{(e^x+1)^6}dx}$$
where $u = (e^y+1), du =e^ydy$ and $v = (e^x+1), dv=e^xdx$
$$int{u^{-2}dy}=-int{v^{-6}}dx$$
$$-u^{-1} + c_1=frac{1}{5}v^{-5}+c_2$$
Now, substitute:
$$-frac{1}{e^y+1}+c_1=frac{1}{5(e^x+1)^5}+c_2$$
The $c$ gets absorbed (also, this is the implicit solution)
$$-frac{1}{e^y+1}=frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
But $-c_3$ is the same as $+c_3$
$$-5(e^x+1)^5+c_3=e^y+1$$
$$-5(e^x+1)^5+c_3=e^y$$
Take the natural log of both sides:
$$ln(-5(e^x+1)^5+c_3)=ln(e^y)$$
$$lnleft(-5(e^x+1)^5+c_3right)=y$$
This is the answer I got. It took long to convert this to Latex, so there might be some discontinuities in the work, but the last answer is for sure what I got.
ordinary-differential-equations
edited Jan 20 at 22:27
El borito
666216
asked Jan 20 at 21:15
Kenneth DangKenneth Dang
136
$endgroup$
This is the problem. It is question 4. I typed the answer in explicit form, but my professor had said it is incorrect. Could someone check over my work, because I think I did have it correct. My Work
Because someone asked for my work due to bad handwriting:
$$(e^y+1)^2e^{-y}dx+(e^x+1)^6e^{-x}dy=0$$
Trying to seperate variables:
$$(e^y+1)^2(e^{-y})dx=-(e^x+1)^6(e^{-x})dy$$
$$frac{-e^x}{(e^x+1)^6}dx=frac{e^y}{(e^y+1)^2}dy$$
$$int{frac{-e^x}{(e^x+1)^6}dx}=int{frac{e^y}{(e^y+1)^2}dy}$$
Flip:
$$int{frac{e^y}{(e^y+1)^2}dy}=int{frac{-e^x}{(e^x+1)^6}dx}$$
where $u = (e^y+1), du =e^ydy$ and $v = (e^x+1), dv=e^xdx$
$$int{u^{-2}dy}=-int{v^{-6}}dx$$
$$-u^{-1} + c_1=frac{1}{5}v^{-5}+c_2$$
Now, substitute:
$$-frac{1}{e^y+1}+c_1=frac{1}{5(e^x+1)^5}+c_2$$
The $c$ gets absorbed (also, this is the implicit solution)
$$-frac{1}{e^y+1}=frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
But $-c_3$ is the same as $+c_3$
$$-5(e^x+1)^5+c_3=e^y+1$$
$$-5(e^x+1)^5+c_3=e^y$$
Take the natural log of both sides:
$$ln(-5(e^x+1)^5+c_3)=ln(e^y)$$
$$lnleft(-5(e^x+1)^5+c_3right)=y$$
This is the answer I got. It took long to convert this to Latex, so there might be some discontinuities in the work, but the last answer is for sure what I got.
ordinary-differential-equations
ordinary-differential-equations
edited Jan 20 at 22:27
El borito
666216
asked Jan 20 at 21:15
Kenneth DangKenneth Dang
136
edited Jan 20 at 22:27
El borito
666216
asked Jan 20 at 21:15
Kenneth DangKenneth Dang
136
edited Jan 20 at 22:27
El borito
666216
edited Jan 20 at 22:27
El borito
666216
edited Jan 20 at 22:27
El borito
666216
666216
asked Jan 20 at 21:15
Kenneth DangKenneth Dang
136
asked Jan 20 at 21:15
Kenneth DangKenneth Dang
136
asked Jan 20 at 21:15
Kenneth DangKenneth Dang
136
136
$begingroup$
You can't be serious!!! Your "work" is impossible to read. How about writing one equation per line? Or better still, how about transcribing your work in TeX?
$endgroup$
– Bernard Massé
Jan 20 at 21:23
$begingroup$
If you think your answer is the right one, just derive and plug it in the equation. If you come up with an equality, then your answer is good. Or use Wolfram Alpha to verify your answer.
$endgroup$
– Bernard Massé
Jan 20 at 21:30
1
$begingroup$
No, $-frac1a=frac1b+c$ is not equivalent to $-b-c=a$. Not even with a different constant.
$endgroup$
– LutzL
Jan 20 at 22:22
1
$begingroup$
Why? Wouldn't that just be $-frac{b}{a}=1-bc$ $implies$ $-frac{b}{a}=1-c$ $implies$ $-frac{b}{a}+c=1$ $implies$ $-b+ac=a$ $implies$ $-b+c=a$? or as you said $-b-c=a$?
$endgroup$
– Kenneth Dang
Jan 20 at 22:29
1
$begingroup$
$a$ and $b$ are the variable terms containing $x$ and $y$, $c$ is the integration constant. Multiplying by $ab$ gives $-b=a+cab$, and the last term is not constant, can thus not be replaced with a new constant.
$endgroup$
– LutzL
Jan 20 at 23:13
|
show 1 more comment
$begingroup$
You can't be serious!!! Your "work" is impossible to read. How about writing one equation per line? Or better still, how about transcribing your work in TeX?
$endgroup$
– Bernard Massé
Jan 20 at 21:23
$begingroup$
If you think your answer is the right one, just derive and plug it in the equation. If you come up with an equality, then your answer is good. Or use Wolfram Alpha to verify your answer.
$endgroup$
– Bernard Massé
Jan 20 at 21:30
1
$begingroup$
No, $-frac1a=frac1b+c$ is not equivalent to $-b-c=a$. Not even with a different constant.
$endgroup$
– LutzL
Jan 20 at 22:22
1
$begingroup$
Why? Wouldn't that just be $-frac{b}{a}=1-bc$ $implies$ $-frac{b}{a}=1-c$ $implies$ $-frac{b}{a}+c=1$ $implies$ $-b+ac=a$ $implies$ $-b+c=a$? or as you said $-b-c=a$?
$endgroup$
– Kenneth Dang
Jan 20 at 22:29
1
$begingroup$
$a$ and $b$ are the variable terms containing $x$ and $y$, $c$ is the integration constant. Multiplying by $ab$ gives $-b=a+cab$, and the last term is not constant, can thus not be replaced with a new constant.
$endgroup$
– LutzL
Jan 20 at 23:13
$begingroup$
You can't be serious!!! Your "work" is impossible to read. How about writing one equation per line? Or better still, how about transcribing your work in TeX?
$endgroup$
– Bernard Massé
Jan 20 at 21:23
$begingroup$
You can't be serious!!! Your "work" is impossible to read. How about writing one equation per line? Or better still, how about transcribing your work in TeX?
$endgroup$
– Bernard Massé
Jan 20 at 21:23
$begingroup$
If you think your answer is the right one, just derive and plug it in the equation. If you come up with an equality, then your answer is good. Or use Wolfram Alpha to verify your answer.
$endgroup$
– Bernard Massé
Jan 20 at 21:30
$begingroup$
If you think your answer is the right one, just derive and plug it in the equation. If you come up with an equality, then your answer is good. Or use Wolfram Alpha to verify your answer.
$endgroup$
– Bernard Massé
Jan 20 at 21:30
1
1
$begingroup$
No, $-frac1a=frac1b+c$ is not equivalent to $-b-c=a$. Not even with a different constant.
$endgroup$
– LutzL
Jan 20 at 22:22
$begingroup$
No, $-frac1a=frac1b+c$ is not equivalent to $-b-c=a$. Not even with a different constant.
$endgroup$
– LutzL
Jan 20 at 22:22
1
1
$begingroup$
Why? Wouldn't that just be $-frac{b}{a}=1-bc$ $implies$ $-frac{b}{a}=1-c$ $implies$ $-frac{b}{a}+c=1$ $implies$ $-b+ac=a$ $implies$ $-b+c=a$? or as you said $-b-c=a$?
$endgroup$
– Kenneth Dang
Jan 20 at 22:29
$begingroup$
Why? Wouldn't that just be $-frac{b}{a}=1-bc$ $implies$ $-frac{b}{a}=1-c$ $implies$ $-frac{b}{a}+c=1$ $implies$ $-b+ac=a$ $implies$ $-b+c=a$? or as you said $-b-c=a$?
$endgroup$
– Kenneth Dang
Jan 20 at 22:29
1
1
$begingroup$
$a$ and $b$ are the variable terms containing $x$ and $y$, $c$ is the integration constant. Multiplying by $ab$ gives $-b=a+cab$, and the last term is not constant, can thus not be replaced with a new constant.
$endgroup$
– LutzL
Jan 20 at 23:13
$begingroup$
$a$ and $b$ are the variable terms containing $x$ and $y$, $c$ is the integration constant. Multiplying by $ab$ gives $-b=a+cab$, and the last term is not constant, can thus not be replaced with a new constant.
$endgroup$
– LutzL
Jan 20 at 23:13
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
The wrong step is here :
$$-frac{1}{e^y+1}=frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
The correct steps are :
Multiply both left-hand and right-hand terms by $(e^y+1)(5(e^x+1)^5)$
$$left(-frac{1}{e^y+1}right)(e^y+1)big(5(e^x+1)^5big)=left(frac{1}{5(e^x+1)^5}+c_3right)(e^y+1)big(5(e^x+1)^5big)$$
Simplify :
$$-big(5(e^x+1)^5big)=left(1+c_3 5(e^x+1)^5right)(e^y+1)$$
$$ e^y+1=frac{-5(e^x+1)^5}{1+ 5c_3(e^x+1)^5} $$
$$y=lnleft(frac{-5(e^x+1)^5}{1+c_3 5(e^x+1)^5}-1 right)$$
edited Jan 21 at 20:17
El borito
666216
answered Jan 21 at 7:49
JJacquelinJJacquelin
44.3k21854
$endgroup$
add a comment |
$begingroup$
The step from:
$$ -frac{1}{e^y + 1} = frac{1}{5left(e^x+1right)^5} +c_1 tag{1} $$
to:
$$ -5(e^x +1)^5-c_2 = e^y +1 tag{2}$$
is wrong. Because if what you say is true then it implies that (1) and (2) are a tautology, but if you replace a term of (2) in (1):
$$ -frac{1}{e^y + 1} = frac{1}{5left(e^x+1right)^5} +c_3 $$
$$ frac{1}{5(e^x +1)^5+c_2} = frac{1}{5left(e^x+1right)^5} +c_3 $$
$$ 1 = frac{5(e^x +1)^5+c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
$$ 1 = 1+frac{c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
$$ 0 = frac{c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
it is a polynomial equation if $u=e^x+1$, and the equation has a finite set of solutions of $u$ (thus $x$), and the set solution isn't a interval in $Bbb{R}$, thus the implication from (1) to (2) is false, becaus it doesn't have sense.
answered Jan 21 at 6:39
El boritoEl borito
666216
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The wrong step is here :
$$-frac{1}{e^y+1}=frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
The correct steps are :
Multiply both left-hand and right-hand terms by $(e^y+1)(5(e^x+1)^5)$
$$left(-frac{1}{e^y+1}right)(e^y+1)big(5(e^x+1)^5big)=left(frac{1}{5(e^x+1)^5}+c_3right)(e^y+1)big(5(e^x+1)^5big)$$
Simplify :
$$-big(5(e^x+1)^5big)=left(1+c_3 5(e^x+1)^5right)(e^y+1)$$
$$ e^y+1=frac{-5(e^x+1)^5}{1+ 5c_3(e^x+1)^5} $$
$$y=lnleft(frac{-5(e^x+1)^5}{1+c_3 5(e^x+1)^5}-1 right)$$
edited Jan 21 at 20:17
El borito
666216
answered Jan 21 at 7:49
JJacquelinJJacquelin
44.3k21854
$endgroup$
add a comment |
$begingroup$
The wrong step is here :
$$-frac{1}{e^y+1}=frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
The correct steps are :
Multiply both left-hand and right-hand terms by $(e^y+1)(5(e^x+1)^5)$
$$left(-frac{1}{e^y+1}right)(e^y+1)big(5(e^x+1)^5big)=left(frac{1}{5(e^x+1)^5}+c_3right)(e^y+1)big(5(e^x+1)^5big)$$
Simplify :
$$-big(5(e^x+1)^5big)=left(1+c_3 5(e^x+1)^5right)(e^y+1)$$
$$ e^y+1=frac{-5(e^x+1)^5}{1+ 5c_3(e^x+1)^5} $$
$$y=lnleft(frac{-5(e^x+1)^5}{1+c_3 5(e^x+1)^5}-1 right)$$
edited Jan 21 at 20:17
El borito
666216
answered Jan 21 at 7:49
JJacquelinJJacquelin
44.3k21854
$endgroup$
add a comment |
$begingroup$
The wrong step is here :
$$-frac{1}{e^y+1}=frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
The correct steps are :
Multiply both left-hand and right-hand terms by $(e^y+1)(5(e^x+1)^5)$
$$left(-frac{1}{e^y+1}right)(e^y+1)big(5(e^x+1)^5big)=left(frac{1}{5(e^x+1)^5}+c_3right)(e^y+1)big(5(e^x+1)^5big)$$
Simplify :
$$-big(5(e^x+1)^5big)=left(1+c_3 5(e^x+1)^5right)(e^y+1)$$
$$ e^y+1=frac{-5(e^x+1)^5}{1+ 5c_3(e^x+1)^5} $$
$$y=lnleft(frac{-5(e^x+1)^5}{1+c_3 5(e^x+1)^5}-1 right)$$
edited Jan 21 at 20:17
El borito
666216
answered Jan 21 at 7:49
JJacquelinJJacquelin
44.3k21854
$endgroup$
The wrong step is here :
$$-frac{1}{e^y+1}=frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
The correct steps are :
Multiply both left-hand and right-hand terms by $(e^y+1)(5(e^x+1)^5)$
$$left(-frac{1}{e^y+1}right)(e^y+1)big(5(e^x+1)^5big)=left(frac{1}{5(e^x+1)^5}+c_3right)(e^y+1)big(5(e^x+1)^5big)$$
Simplify :
$$-big(5(e^x+1)^5big)=left(1+c_3 5(e^x+1)^5right)(e^y+1)$$
$$ e^y+1=frac{-5(e^x+1)^5}{1+ 5c_3(e^x+1)^5} $$
$$y=lnleft(frac{-5(e^x+1)^5}{1+c_3 5(e^x+1)^5}-1 right)$$
edited Jan 21 at 20:17
El borito
666216
answered Jan 21 at 7:49
JJacquelinJJacquelin
44.3k21854
edited Jan 21 at 20:17
El borito
666216
edited Jan 21 at 20:17
El borito
666216
edited Jan 21 at 20:17
El borito
666216
666216
answered Jan 21 at 7:49
JJacquelinJJacquelin
44.3k21854
answered Jan 21 at 7:49
JJacquelinJJacquelin
44.3k21854
answered Jan 21 at 7:49
JJacquelinJJacquelin
44.3k21854
44.3k21854
add a comment |
add a comment |
$begingroup$
The step from:
$$ -frac{1}{e^y + 1} = frac{1}{5left(e^x+1right)^5} +c_1 tag{1} $$
to:
$$ -5(e^x +1)^5-c_2 = e^y +1 tag{2}$$
is wrong. Because if what you say is true then it implies that (1) and (2) are a tautology, but if you replace a term of (2) in (1):
$$ -frac{1}{e^y + 1} = frac{1}{5left(e^x+1right)^5} +c_3 $$
$$ frac{1}{5(e^x +1)^5+c_2} = frac{1}{5left(e^x+1right)^5} +c_3 $$
$$ 1 = frac{5(e^x +1)^5+c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
$$ 1 = 1+frac{c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
$$ 0 = frac{c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
it is a polynomial equation if $u=e^x+1$, and the equation has a finite set of solutions of $u$ (thus $x$), and the set solution isn't a interval in $Bbb{R}$, thus the implication from (1) to (2) is false, becaus it doesn't have sense.
answered Jan 21 at 6:39
El boritoEl borito
666216
$endgroup$
add a comment |
$begingroup$
The step from:
$$ -frac{1}{e^y + 1} = frac{1}{5left(e^x+1right)^5} +c_1 tag{1} $$
to:
$$ -5(e^x +1)^5-c_2 = e^y +1 tag{2}$$
is wrong. Because if what you say is true then it implies that (1) and (2) are a tautology, but if you replace a term of (2) in (1):
$$ -frac{1}{e^y + 1} = frac{1}{5left(e^x+1right)^5} +c_3 $$
$$ frac{1}{5(e^x +1)^5+c_2} = frac{1}{5left(e^x+1right)^5} +c_3 $$
$$ 1 = frac{5(e^x +1)^5+c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
$$ 1 = 1+frac{c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
$$ 0 = frac{c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
it is a polynomial equation if $u=e^x+1$, and the equation has a finite set of solutions of $u$ (thus $x$), and the set solution isn't a interval in $Bbb{R}$, thus the implication from (1) to (2) is false, becaus it doesn't have sense.
answered Jan 21 at 6:39
El boritoEl borito
666216
$endgroup$
add a comment |
$begingroup$
The step from:
$$ -frac{1}{e^y + 1} = frac{1}{5left(e^x+1right)^5} +c_1 tag{1} $$
to:
$$ -5(e^x +1)^5-c_2 = e^y +1 tag{2}$$
is wrong. Because if what you say is true then it implies that (1) and (2) are a tautology, but if you replace a term of (2) in (1):
$$ -frac{1}{e^y + 1} = frac{1}{5left(e^x+1right)^5} +c_3 $$
$$ frac{1}{5(e^x +1)^5+c_2} = frac{1}{5left(e^x+1right)^5} +c_3 $$
$$ 1 = frac{5(e^x +1)^5+c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
$$ 1 = 1+frac{c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
$$ 0 = frac{c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
it is a polynomial equation if $u=e^x+1$, and the equation has a finite set of solutions of $u$ (thus $x$), and the set solution isn't a interval in $Bbb{R}$, thus the implication from (1) to (2) is false, becaus it doesn't have sense.
answered Jan 21 at 6:39
El boritoEl borito
666216
$endgroup$
The step from:
$$ -frac{1}{e^y + 1} = frac{1}{5left(e^x+1right)^5} +c_1 tag{1} $$
to:
$$ -5(e^x +1)^5-c_2 = e^y +1 tag{2}$$
is wrong. Because if what you say is true then it implies that (1) and (2) are a tautology, but if you replace a term of (2) in (1):
$$ -frac{1}{e^y + 1} = frac{1}{5left(e^x+1right)^5} +c_3 $$
$$ frac{1}{5(e^x +1)^5+c_2} = frac{1}{5left(e^x+1right)^5} +c_3 $$
$$ 1 = frac{5(e^x +1)^5+c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
$$ 1 = 1+frac{c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
$$ 0 = frac{c_2}{5left(e^x+1right)^5} +c_3big[5(e^x +1)^5+c_2big] $$
it is a polynomial equation if $u=e^x+1$, and the equation has a finite set of solutions of $u$ (thus $x$), and the set solution isn't a interval in $Bbb{R}$, thus the implication from (1) to (2) is false, becaus it doesn't have sense.
answered Jan 21 at 6:39
El boritoEl borito
666216
answered Jan 21 at 6:39
El boritoEl borito
666216
answered Jan 21 at 6:39
El boritoEl borito
666216
answered Jan 21 at 6:39
El boritoEl borito
666216
666216
add a comment |
add a comment |
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$begingroup$
You can't be serious!!! Your "work" is impossible to read. How about writing one equation per line? Or better still, how about transcribing your work in TeX?
$endgroup$
– Bernard Massé
Jan 20 at 21:23
$begingroup$
If you think your answer is the right one, just derive and plug it in the equation. If you come up with an equality, then your answer is good. Or use Wolfram Alpha to verify your answer.
$endgroup$
– Bernard Massé
Jan 20 at 21:30
1
$begingroup$
No, $-frac1a=frac1b+c$ is not equivalent to $-b-c=a$. Not even with a different constant.
$endgroup$
– LutzL
Jan 20 at 22:22
1
$begingroup$
Why? Wouldn't that just be $-frac{b}{a}=1-bc$ $implies$ $-frac{b}{a}=1-c$ $implies$ $-frac{b}{a}+c=1$ $implies$ $-b+ac=a$ $implies$ $-b+c=a$? or as you said $-b-c=a$?
$endgroup$
– Kenneth Dang
Jan 20 at 22:29
1
$begingroup$
$a$ and $b$ are the variable terms containing $x$ and $y$, $c$ is the integration constant. Multiplying by $ab$ gives $-b=a+cab$, and the last term is not constant, can thus not be replaced with a new constant.
$endgroup$
– LutzL
Jan 20 at 23:13