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







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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







    share|cite|improve this question


























      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







      share|cite|improve this question















      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






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      edited Nov 28 at 21:17

























      asked Nov 28 at 20:48









      user10478

      376111




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          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).






          share|cite|improve this answer




























            up vote
            4
            down vote













            Your solution is wrong as
            $$
            intfrac{du}{1+u^2}=arctan(u),
            $$
            so that
            $$
            u=tan(t+c),~~ y=tan(t+c)-t.
            $$






            share|cite|improve this answer

















            • 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











            Your Answer





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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).






            share|cite|improve this answer

























              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).






              share|cite|improve this answer























                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).






                share|cite|improve this answer












                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).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 4 at 0:49









                Barry Cipra

                58.6k653122




                58.6k653122






















                    up vote
                    4
                    down vote













                    Your solution is wrong as
                    $$
                    intfrac{du}{1+u^2}=arctan(u),
                    $$
                    so that
                    $$
                    u=tan(t+c),~~ y=tan(t+c)-t.
                    $$






                    share|cite|improve this answer

















                    • 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















                    up vote
                    4
                    down vote













                    Your solution is wrong as
                    $$
                    intfrac{du}{1+u^2}=arctan(u),
                    $$
                    so that
                    $$
                    u=tan(t+c),~~ y=tan(t+c)-t.
                    $$






                    share|cite|improve this answer

















                    • 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













                    up vote
                    4
                    down vote










                    up vote
                    4
                    down vote









                    Your solution is wrong as
                    $$
                    intfrac{du}{1+u^2}=arctan(u),
                    $$
                    so that
                    $$
                    u=tan(t+c),~~ y=tan(t+c)-t.
                    $$






                    share|cite|improve this answer












                    Your solution is wrong as
                    $$
                    intfrac{du}{1+u^2}=arctan(u),
                    $$
                    so that
                    $$
                    u=tan(t+c),~~ y=tan(t+c)-t.
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    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














                    • 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


















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