Quadratic Formula With Independent and Dependent Variables
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Given the differential equation $dy/dt = (y + t)^2$, we can apply the u-substitution $u = y + t$ to arrive at the separable differential equation $du/dt = u^2 + 1$. This separates to $1/(u^2 + 1) du = dt$ which integrates to (EDIT: As LutzL has pointed out, I integrated incorrectly. However, correcting it would eclipse potentially interesting part of the question, so I'll leave the mistake) $u^2 + 1 - Ce^t = 0$. Reverting the substitution yields $y^2 + 2ty + t^2 + 1 - Ce^t = 0$. Note that $y$ is a dependent variable, $t$ is the independent variable, and $C$ is an arbitrary constant.
Is it legal to proceed via the quadratic formula, using the appropriate expressions in terms of $t$ as coefficients? This would look like $y = (-(2t) ± sqrt{(2t)^2 - 4(1)(t^2 + 1 - Ce^t)}) / 2(1)$, which works out to $y = -t ± sqrt{Ce^t - 1}$. However, this practice feels a bit suspect, since in other instances of applying the quadratic formula, there is no dependency between the variable and its coefficients, whereas here there is. Is this a legal and correct approach to the problem?
Secondly, suppose a similar problem yielded $y^2 + 2ty + t^2 + 1 − Ce^y = 0$, where $y$ is still a dependent variable, $t$ is still the independent variable, and $C$ is still an arbitrary constant. Would it be legal to solve for $t$ using the quadratic formula using the appropriate expressions in terms of $y$ as coefficients?
differential-equations independence constants
asked Nov 28 at 20:48
user10478
376111
add a comment |
up vote
4
down vote
favorite
Given the differential equation $dy/dt = (y + t)^2$, we can apply the u-substitution $u = y + t$ to arrive at the separable differential equation $du/dt = u^2 + 1$. This separates to $1/(u^2 + 1) du = dt$ which integrates to (EDIT: As LutzL has pointed out, I integrated incorrectly. However, correcting it would eclipse potentially interesting part of the question, so I'll leave the mistake) $u^2 + 1 - Ce^t = 0$. Reverting the substitution yields $y^2 + 2ty + t^2 + 1 - Ce^t = 0$. Note that $y$ is a dependent variable, $t$ is the independent variable, and $C$ is an arbitrary constant.
Is it legal to proceed via the quadratic formula, using the appropriate expressions in terms of $t$ as coefficients? This would look like $y = (-(2t) ± sqrt{(2t)^2 - 4(1)(t^2 + 1 - Ce^t)}) / 2(1)$, which works out to $y = -t ± sqrt{Ce^t - 1}$. However, this practice feels a bit suspect, since in other instances of applying the quadratic formula, there is no dependency between the variable and its coefficients, whereas here there is. Is this a legal and correct approach to the problem?
Secondly, suppose a similar problem yielded $y^2 + 2ty + t^2 + 1 − Ce^y = 0$, where $y$ is still a dependent variable, $t$ is still the independent variable, and $C$ is still an arbitrary constant. Would it be legal to solve for $t$ using the quadratic formula using the appropriate expressions in terms of $y$ as coefficients?
differential-equations independence constants
asked Nov 28 at 20:48
user10478
376111
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Given the differential equation $dy/dt = (y + t)^2$, we can apply the u-substitution $u = y + t$ to arrive at the separable differential equation $du/dt = u^2 + 1$. This separates to $1/(u^2 + 1) du = dt$ which integrates to (EDIT: As LutzL has pointed out, I integrated incorrectly. However, correcting it would eclipse potentially interesting part of the question, so I'll leave the mistake) $u^2 + 1 - Ce^t = 0$. Reverting the substitution yields $y^2 + 2ty + t^2 + 1 - Ce^t = 0$. Note that $y$ is a dependent variable, $t$ is the independent variable, and $C$ is an arbitrary constant.
Is it legal to proceed via the quadratic formula, using the appropriate expressions in terms of $t$ as coefficients? This would look like $y = (-(2t) ± sqrt{(2t)^2 - 4(1)(t^2 + 1 - Ce^t)}) / 2(1)$, which works out to $y = -t ± sqrt{Ce^t - 1}$. However, this practice feels a bit suspect, since in other instances of applying the quadratic formula, there is no dependency between the variable and its coefficients, whereas here there is. Is this a legal and correct approach to the problem?
Secondly, suppose a similar problem yielded $y^2 + 2ty + t^2 + 1 − Ce^y = 0$, where $y$ is still a dependent variable, $t$ is still the independent variable, and $C$ is still an arbitrary constant. Would it be legal to solve for $t$ using the quadratic formula using the appropriate expressions in terms of $y$ as coefficients?
differential-equations independence constants
asked Nov 28 at 20:48
user10478
376111
Given the differential equation $dy/dt = (y + t)^2$, we can apply the u-substitution $u = y + t$ to arrive at the separable differential equation $du/dt = u^2 + 1$. This separates to $1/(u^2 + 1) du = dt$ which integrates to (EDIT: As LutzL has pointed out, I integrated incorrectly. However, correcting it would eclipse potentially interesting part of the question, so I'll leave the mistake) $u^2 + 1 - Ce^t = 0$. Reverting the substitution yields $y^2 + 2ty + t^2 + 1 - Ce^t = 0$. Note that $y$ is a dependent variable, $t$ is the independent variable, and $C$ is an arbitrary constant.
Is it legal to proceed via the quadratic formula, using the appropriate expressions in terms of $t$ as coefficients? This would look like $y = (-(2t) ± sqrt{(2t)^2 - 4(1)(t^2 + 1 - Ce^t)}) / 2(1)$, which works out to $y = -t ± sqrt{Ce^t - 1}$. However, this practice feels a bit suspect, since in other instances of applying the quadratic formula, there is no dependency between the variable and its coefficients, whereas here there is. Is this a legal and correct approach to the problem?
Secondly, suppose a similar problem yielded $y^2 + 2ty + t^2 + 1 − Ce^y = 0$, where $y$ is still a dependent variable, $t$ is still the independent variable, and $C$ is still an arbitrary constant. Would it be legal to solve for $t$ using the quadratic formula using the appropriate expressions in terms of $y$ as coefficients?
differential-equations independence constants
differential-equations independence constants
asked Nov 28 at 20:48
user10478
376111
asked Nov 28 at 20:48
user10478
376111
edited Nov 28 at 21:17
asked Nov 28 at 20:48
user10478
376111
asked Nov 28 at 20:48
user10478
376111
asked Nov 28 at 20:48
user10478
376111
376111
add a comment |
add a comment |
2 Answers
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User Lutzl has addressed an error in the set-up of your question. But in answer to the question itself, yes, you can apply the quadratic formula anytime you have a quadratic expression, so $y^2+2ty+t^2+1-Ce^t=0$ implies $y=-tpmsqrt{Ce^t-1}$ and $y^2+2ty+t^2+1-Ce^y=0$ implies $t=-ypmsqrt{Ce^y-1}$ (the key difference being that you're hard pressed to invert the expression $-ypmsqrt{Ce^y-1}$ to get an explicit formula for $y$ as a function of $t$). For that matter, it's even OK (if pointless) to say
$$y^2+2ty+t^2+1-Ce^y=0implies y=-tpmsqrt{Ce^y-1}$$
All you're really doing in any of these is saying
$$y^2+2ty+t^2=whateverimplies (y+t)^2=whateverimplies y+t=pmsqrt{whatever}$$
(where that whatever better be nonnegative, unless you're prepared to deal with complex numbers).
answered Dec 4 at 0:49
Barry Cipra
58.6k653122
add a comment |
up vote
4
down vote
Your solution is wrong as
$$
intfrac{du}{1+u^2}=arctan(u),
$$
so that
$$
u=tan(t+c),~~ y=tan(t+c)-t.
$$
answered Nov 28 at 21:09
LutzL
54.8k42053
1
Darn, thanks for catching that. :P I'm going to leave the mistake as is with a note, as correcting it would be harmful to the rest of the question.
– user10478
Nov 28 at 21:19
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
up vote
3
down vote
accepted
User Lutzl has addressed an error in the set-up of your question. But in answer to the question itself, yes, you can apply the quadratic formula anytime you have a quadratic expression, so $y^2+2ty+t^2+1-Ce^t=0$ implies $y=-tpmsqrt{Ce^t-1}$ and $y^2+2ty+t^2+1-Ce^y=0$ implies $t=-ypmsqrt{Ce^y-1}$ (the key difference being that you're hard pressed to invert the expression $-ypmsqrt{Ce^y-1}$ to get an explicit formula for $y$ as a function of $t$). For that matter, it's even OK (if pointless) to say
$$y^2+2ty+t^2+1-Ce^y=0implies y=-tpmsqrt{Ce^y-1}$$
All you're really doing in any of these is saying
$$y^2+2ty+t^2=whateverimplies (y+t)^2=whateverimplies y+t=pmsqrt{whatever}$$
(where that whatever better be nonnegative, unless you're prepared to deal with complex numbers).
answered Dec 4 at 0:49
Barry Cipra
58.6k653122
add a comment |
up vote
3
down vote
accepted
User Lutzl has addressed an error in the set-up of your question. But in answer to the question itself, yes, you can apply the quadratic formula anytime you have a quadratic expression, so $y^2+2ty+t^2+1-Ce^t=0$ implies $y=-tpmsqrt{Ce^t-1}$ and $y^2+2ty+t^2+1-Ce^y=0$ implies $t=-ypmsqrt{Ce^y-1}$ (the key difference being that you're hard pressed to invert the expression $-ypmsqrt{Ce^y-1}$ to get an explicit formula for $y$ as a function of $t$). For that matter, it's even OK (if pointless) to say
$$y^2+2ty+t^2+1-Ce^y=0implies y=-tpmsqrt{Ce^y-1}$$
All you're really doing in any of these is saying
$$y^2+2ty+t^2=whateverimplies (y+t)^2=whateverimplies y+t=pmsqrt{whatever}$$
(where that whatever better be nonnegative, unless you're prepared to deal with complex numbers).
answered Dec 4 at 0:49
Barry Cipra
58.6k653122
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
User Lutzl has addressed an error in the set-up of your question. But in answer to the question itself, yes, you can apply the quadratic formula anytime you have a quadratic expression, so $y^2+2ty+t^2+1-Ce^t=0$ implies $y=-tpmsqrt{Ce^t-1}$ and $y^2+2ty+t^2+1-Ce^y=0$ implies $t=-ypmsqrt{Ce^y-1}$ (the key difference being that you're hard pressed to invert the expression $-ypmsqrt{Ce^y-1}$ to get an explicit formula for $y$ as a function of $t$). For that matter, it's even OK (if pointless) to say
$$y^2+2ty+t^2+1-Ce^y=0implies y=-tpmsqrt{Ce^y-1}$$
All you're really doing in any of these is saying
$$y^2+2ty+t^2=whateverimplies (y+t)^2=whateverimplies y+t=pmsqrt{whatever}$$
(where that whatever better be nonnegative, unless you're prepared to deal with complex numbers).
answered Dec 4 at 0:49
Barry Cipra
58.6k653122
User Lutzl has addressed an error in the set-up of your question. But in answer to the question itself, yes, you can apply the quadratic formula anytime you have a quadratic expression, so $y^2+2ty+t^2+1-Ce^t=0$ implies $y=-tpmsqrt{Ce^t-1}$ and $y^2+2ty+t^2+1-Ce^y=0$ implies $t=-ypmsqrt{Ce^y-1}$ (the key difference being that you're hard pressed to invert the expression $-ypmsqrt{Ce^y-1}$ to get an explicit formula for $y$ as a function of $t$). For that matter, it's even OK (if pointless) to say
$$y^2+2ty+t^2+1-Ce^y=0implies y=-tpmsqrt{Ce^y-1}$$
All you're really doing in any of these is saying
$$y^2+2ty+t^2=whateverimplies (y+t)^2=whateverimplies y+t=pmsqrt{whatever}$$
(where that whatever better be nonnegative, unless you're prepared to deal with complex numbers).
answered Dec 4 at 0:49
Barry Cipra
58.6k653122
answered Dec 4 at 0:49
Barry Cipra
58.6k653122
answered Dec 4 at 0:49
Barry Cipra
58.6k653122
answered Dec 4 at 0:49
Barry Cipra
58.6k653122
58.6k653122
add a comment |
add a comment |
up vote
4
down vote
Your solution is wrong as
$$
intfrac{du}{1+u^2}=arctan(u),
$$
so that
$$
u=tan(t+c),~~ y=tan(t+c)-t.
$$
answered Nov 28 at 21:09
LutzL
54.8k42053
1
Darn, thanks for catching that. :P I'm going to leave the mistake as is with a note, as correcting it would be harmful to the rest of the question.
– user10478
Nov 28 at 21:19
add a comment |
up vote
4
down vote
Your solution is wrong as
$$
intfrac{du}{1+u^2}=arctan(u),
$$
so that
$$
u=tan(t+c),~~ y=tan(t+c)-t.
$$
answered Nov 28 at 21:09
LutzL
54.8k42053
1
Darn, thanks for catching that. :P I'm going to leave the mistake as is with a note, as correcting it would be harmful to the rest of the question.
– user10478
Nov 28 at 21:19
add a comment |
up vote
4
down vote
up vote
4
down vote
Your solution is wrong as
$$
intfrac{du}{1+u^2}=arctan(u),
$$
so that
$$
u=tan(t+c),~~ y=tan(t+c)-t.
$$
answered Nov 28 at 21:09
LutzL
54.8k42053
Your solution is wrong as
$$
intfrac{du}{1+u^2}=arctan(u),
$$
so that
$$
u=tan(t+c),~~ y=tan(t+c)-t.
$$
answered Nov 28 at 21:09
LutzL
54.8k42053
answered Nov 28 at 21:09
LutzL
54.8k42053
answered Nov 28 at 21:09
LutzL
54.8k42053
answered Nov 28 at 21:09
LutzL
54.8k42053
54.8k42053
1
Darn, thanks for catching that. :P I'm going to leave the mistake as is with a note, as correcting it would be harmful to the rest of the question.
– user10478
Nov 28 at 21:19
add a comment |
1
Darn, thanks for catching that. :P I'm going to leave the mistake as is with a note, as correcting it would be harmful to the rest of the question.
– user10478
Nov 28 at 21:19
1
1
Darn, thanks for catching that. :P I'm going to leave the mistake as is with a note, as correcting it would be harmful to the rest of the question.
– user10478
Nov 28 at 21:19
Darn, thanks for catching that. :P I'm going to leave the mistake as is with a note, as correcting it would be harmful to the rest of the question.
– user10478
Nov 28 at 21:19
add a comment |
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