Does the constant $C$ in this solution to a differential equation equal infinity?












2












$begingroup$


The problem is $y' = -frac{1}{t^2} - frac{1}{t}y + y^2; y_p = frac{1}{t}$. My solution is



$$begin{align}
y = frac{1}{t} + B &implies y' = -frac{1}{t^2} + B' \
&implies -frac{1}{t^2} - frac{1}{t}y + y^2 = -frac{1}{t^2} + B' \
&implies -frac{1}{t^2} - frac{1}{t} left(frac{1}{t} + B right) + left(frac{1}{t} + Bright)^2 = -frac{1}{t^2} + B' \
&implies B' - frac{1}{t}B = B^2 \
&implies L = B^{-1} \
&implies L' = -B^{-2} left(B^2 + frac{1}{t}B right) \
&implies L' + frac{1}{tB} = -1 \
&implies L' + frac{1}{t}L = -1 \
&implies L_h = frac{1}{t} \
&implies L = frac{1}{t}intfrac{-1}{frac{1}{t}}dt \
&implies L = frac{1}{t} left(-frac{1}{2}t^2 + C_{tentative} right) \
&implies L = frac{C - t^2}{2t} \
&implies B = frac{2t}{C - t^2} \
&implies y = frac{1}{t} + frac{2t}{C - t^2}
end{align}$$



Although this solution does not appear equivalent to the answer given in my book, one or two people in chat reviewed this work and could not see anything incorrect. The issue is that the given particular solution, $y_p = frac{1}{t}$, cannot be obtained by plugging any finite value into the $C$ in my general solution. We could say that the general solution yields the given particular solution when $C =$ infinity, more specifically, when $C =$ some $aleph$ expression which makes the term $frac{2t}{C - t^2}$ disappear regardless of $t$, but this feels a bit outside the box for a textbook problem. Is my general solution correct, and if so, how if at all can we derive the given particular solution from it?






ordinary-differential-equations proof-verification infinity substitution constants







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Your solution is not the "general solution" in the sense that it produces all possible solutions. In your deduction, when you made $L=B^{-1}$, you implicitly excluded $B=0$. That is why you can not obtain $y_p$ from your "general" solution.
    $endgroup$
    – Ramiro
    Dec 31 '18 at 5:47










  • $begingroup$
    Is there a way to test the B = 0 case, as there is when dividing an algebraic equation by a variable?
    $endgroup$
    – user10478
    Dec 31 '18 at 15:59










  • $begingroup$
    The most general form would be to consider $C= a/b$, then we get: $$ y = frac{1}{t} + frac{2bt}{a - bt^2} $$ which covers the solution $y_p$ (set $b=0$ and $a neq 0$) and also the solution $ y = -frac{1}{t}$ (set $a=0$ and $bneq 0$), and, of course, all the other solutions.
    $endgroup$
    – Ramiro
    Jan 1 at 15:30


















2












$begingroup$


The problem is $y' = -frac{1}{t^2} - frac{1}{t}y + y^2; y_p = frac{1}{t}$. My solution is



$$begin{align}
y = frac{1}{t} + B &implies y' = -frac{1}{t^2} + B' \
&implies -frac{1}{t^2} - frac{1}{t}y + y^2 = -frac{1}{t^2} + B' \
&implies -frac{1}{t^2} - frac{1}{t} left(frac{1}{t} + B right) + left(frac{1}{t} + Bright)^2 = -frac{1}{t^2} + B' \
&implies B' - frac{1}{t}B = B^2 \
&implies L = B^{-1} \
&implies L' = -B^{-2} left(B^2 + frac{1}{t}B right) \
&implies L' + frac{1}{tB} = -1 \
&implies L' + frac{1}{t}L = -1 \
&implies L_h = frac{1}{t} \
&implies L = frac{1}{t}intfrac{-1}{frac{1}{t}}dt \
&implies L = frac{1}{t} left(-frac{1}{2}t^2 + C_{tentative} right) \
&implies L = frac{C - t^2}{2t} \
&implies B = frac{2t}{C - t^2} \
&implies y = frac{1}{t} + frac{2t}{C - t^2}
end{align}$$



Although this solution does not appear equivalent to the answer given in my book, one or two people in chat reviewed this work and could not see anything incorrect. The issue is that the given particular solution, $y_p = frac{1}{t}$, cannot be obtained by plugging any finite value into the $C$ in my general solution. We could say that the general solution yields the given particular solution when $C =$ infinity, more specifically, when $C =$ some $aleph$ expression which makes the term $frac{2t}{C - t^2}$ disappear regardless of $t$, but this feels a bit outside the box for a textbook problem. Is my general solution correct, and if so, how if at all can we derive the given particular solution from it?






ordinary-differential-equations proof-verification infinity substitution constants







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Your solution is not the "general solution" in the sense that it produces all possible solutions. In your deduction, when you made $L=B^{-1}$, you implicitly excluded $B=0$. That is why you can not obtain $y_p$ from your "general" solution.
    $endgroup$
    – Ramiro
    Dec 31 '18 at 5:47










  • $begingroup$
    Is there a way to test the B = 0 case, as there is when dividing an algebraic equation by a variable?
    $endgroup$
    – user10478
    Dec 31 '18 at 15:59










  • $begingroup$
    The most general form would be to consider $C= a/b$, then we get: $$ y = frac{1}{t} + frac{2bt}{a - bt^2} $$ which covers the solution $y_p$ (set $b=0$ and $a neq 0$) and also the solution $ y = -frac{1}{t}$ (set $a=0$ and $bneq 0$), and, of course, all the other solutions.
    $endgroup$
    – Ramiro
    Jan 1 at 15:30
















2












2








2





$begingroup$


The problem is $y' = -frac{1}{t^2} - frac{1}{t}y + y^2; y_p = frac{1}{t}$. My solution is



$$begin{align}
y = frac{1}{t} + B &implies y' = -frac{1}{t^2} + B' \
&implies -frac{1}{t^2} - frac{1}{t}y + y^2 = -frac{1}{t^2} + B' \
&implies -frac{1}{t^2} - frac{1}{t} left(frac{1}{t} + B right) + left(frac{1}{t} + Bright)^2 = -frac{1}{t^2} + B' \
&implies B' - frac{1}{t}B = B^2 \
&implies L = B^{-1} \
&implies L' = -B^{-2} left(B^2 + frac{1}{t}B right) \
&implies L' + frac{1}{tB} = -1 \
&implies L' + frac{1}{t}L = -1 \
&implies L_h = frac{1}{t} \
&implies L = frac{1}{t}intfrac{-1}{frac{1}{t}}dt \
&implies L = frac{1}{t} left(-frac{1}{2}t^2 + C_{tentative} right) \
&implies L = frac{C - t^2}{2t} \
&implies B = frac{2t}{C - t^2} \
&implies y = frac{1}{t} + frac{2t}{C - t^2}
end{align}$$



Although this solution does not appear equivalent to the answer given in my book, one or two people in chat reviewed this work and could not see anything incorrect. The issue is that the given particular solution, $y_p = frac{1}{t}$, cannot be obtained by plugging any finite value into the $C$ in my general solution. We could say that the general solution yields the given particular solution when $C =$ infinity, more specifically, when $C =$ some $aleph$ expression which makes the term $frac{2t}{C - t^2}$ disappear regardless of $t$, but this feels a bit outside the box for a textbook problem. Is my general solution correct, and if so, how if at all can we derive the given particular solution from it?






ordinary-differential-equations proof-verification infinity substitution constants







share|cite|improve this question











$endgroup$




The problem is $y' = -frac{1}{t^2} - frac{1}{t}y + y^2; y_p = frac{1}{t}$. My solution is



$$begin{align}
y = frac{1}{t} + B &implies y' = -frac{1}{t^2} + B' \
&implies -frac{1}{t^2} - frac{1}{t}y + y^2 = -frac{1}{t^2} + B' \
&implies -frac{1}{t^2} - frac{1}{t} left(frac{1}{t} + B right) + left(frac{1}{t} + Bright)^2 = -frac{1}{t^2} + B' \
&implies B' - frac{1}{t}B = B^2 \
&implies L = B^{-1} \
&implies L' = -B^{-2} left(B^2 + frac{1}{t}B right) \
&implies L' + frac{1}{tB} = -1 \
&implies L' + frac{1}{t}L = -1 \
&implies L_h = frac{1}{t} \
&implies L = frac{1}{t}intfrac{-1}{frac{1}{t}}dt \
&implies L = frac{1}{t} left(-frac{1}{2}t^2 + C_{tentative} right) \
&implies L = frac{C - t^2}{2t} \
&implies B = frac{2t}{C - t^2} \
&implies y = frac{1}{t} + frac{2t}{C - t^2}
end{align}$$



Although this solution does not appear equivalent to the answer given in my book, one or two people in chat reviewed this work and could not see anything incorrect. The issue is that the given particular solution, $y_p = frac{1}{t}$, cannot be obtained by plugging any finite value into the $C$ in my general solution. We could say that the general solution yields the given particular solution when $C =$ infinity, more specifically, when $C =$ some $aleph$ expression which makes the term $frac{2t}{C - t^2}$ disappear regardless of $t$, but this feels a bit outside the box for a textbook problem. Is my general solution correct, and if so, how if at all can we derive the given particular solution from it?






ordinary-differential-equations proof-verification infinity substitution constants




ordinary-differential-equations proof-verification infinity substitution constants






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 5:18









Eevee Trainer

5,4691936




5,4691936










asked Dec 31 '18 at 5:16









user10478user10478

436211




436211








  • 1




    $begingroup$
    Your solution is not the "general solution" in the sense that it produces all possible solutions. In your deduction, when you made $L=B^{-1}$, you implicitly excluded $B=0$. That is why you can not obtain $y_p$ from your "general" solution.
    $endgroup$
    – Ramiro
    Dec 31 '18 at 5:47










  • $begingroup$
    Is there a way to test the B = 0 case, as there is when dividing an algebraic equation by a variable?
    $endgroup$
    – user10478
    Dec 31 '18 at 15:59










  • $begingroup$
    The most general form would be to consider $C= a/b$, then we get: $$ y = frac{1}{t} + frac{2bt}{a - bt^2} $$ which covers the solution $y_p$ (set $b=0$ and $a neq 0$) and also the solution $ y = -frac{1}{t}$ (set $a=0$ and $bneq 0$), and, of course, all the other solutions.
    $endgroup$
    – Ramiro
    Jan 1 at 15:30
















  • 1




    $begingroup$
    Your solution is not the "general solution" in the sense that it produces all possible solutions. In your deduction, when you made $L=B^{-1}$, you implicitly excluded $B=0$. That is why you can not obtain $y_p$ from your "general" solution.
    $endgroup$
    – Ramiro
    Dec 31 '18 at 5:47










  • $begingroup$
    Is there a way to test the B = 0 case, as there is when dividing an algebraic equation by a variable?
    $endgroup$
    – user10478
    Dec 31 '18 at 15:59










  • $begingroup$
    The most general form would be to consider $C= a/b$, then we get: $$ y = frac{1}{t} + frac{2bt}{a - bt^2} $$ which covers the solution $y_p$ (set $b=0$ and $a neq 0$) and also the solution $ y = -frac{1}{t}$ (set $a=0$ and $bneq 0$), and, of course, all the other solutions.
    $endgroup$
    – Ramiro
    Jan 1 at 15:30










1




1




$begingroup$
Your solution is not the "general solution" in the sense that it produces all possible solutions. In your deduction, when you made $L=B^{-1}$, you implicitly excluded $B=0$. That is why you can not obtain $y_p$ from your "general" solution.
$endgroup$
– Ramiro
Dec 31 '18 at 5:47




$begingroup$
Your solution is not the "general solution" in the sense that it produces all possible solutions. In your deduction, when you made $L=B^{-1}$, you implicitly excluded $B=0$. That is why you can not obtain $y_p$ from your "general" solution.
$endgroup$
– Ramiro
Dec 31 '18 at 5:47












$begingroup$
Is there a way to test the B = 0 case, as there is when dividing an algebraic equation by a variable?
$endgroup$
– user10478
Dec 31 '18 at 15:59




$begingroup$
Is there a way to test the B = 0 case, as there is when dividing an algebraic equation by a variable?
$endgroup$
– user10478
Dec 31 '18 at 15:59












$begingroup$
The most general form would be to consider $C= a/b$, then we get: $$ y = frac{1}{t} + frac{2bt}{a - bt^2} $$ which covers the solution $y_p$ (set $b=0$ and $a neq 0$) and also the solution $ y = -frac{1}{t}$ (set $a=0$ and $bneq 0$), and, of course, all the other solutions.
$endgroup$
– Ramiro
Jan 1 at 15:30






$begingroup$
The most general form would be to consider $C= a/b$, then we get: $$ y = frac{1}{t} + frac{2bt}{a - bt^2} $$ which covers the solution $y_p$ (set $b=0$ and $a neq 0$) and also the solution $ y = -frac{1}{t}$ (set $a=0$ and $bneq 0$), and, of course, all the other solutions.
$endgroup$
– Ramiro
Jan 1 at 15:30












2 Answers
2






active

oldest

votes


















5












$begingroup$

This is a very common situation. A general solution for a nonlinear $n$-th order differential equation is a solution depending on $n$ independent parameters, but it does not necessarily give you all the solutions. As in this case, some may be obtained as limits where some parameters go to $infty$. There are also cases where some particular solutions (singular solutions) are obtained as envelopes of the general solution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does the general solution always give the "main" families of solutions? For example, having graphed a general solution on Desmos with a slider representing each integration constant, letting one slider play at a time produces a smoothly transitioning animated sequence of curves. Is this reliable as a means of seeing all particular solutions except outliers, or can there be entire animated sequences not captured by the general solution?
    $endgroup$
    – user10478
    Dec 31 '18 at 16:11



















1












$begingroup$

For non-linear equations like this, it is valid to take $Ctoinfty$. You can also write $c = frac{1}{C}$ to get an alternate form



$$ y(t) = frac{1}{t} + frac{2ct}{1-ct^2} $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would we actually have to think of $C$ as some sort of $aleph$, an actual infinite quantity larger than any real number, rather than a finite real approaching $infty$? If we use the usual calculus notion of approaching infinity, it seems we could always find a $|t|$ large enough relative to $C$ that the term doesn't disappear.
    $endgroup$
    – user10478
    Dec 31 '18 at 16:20













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

This is a very common situation. A general solution for a nonlinear $n$-th order differential equation is a solution depending on $n$ independent parameters, but it does not necessarily give you all the solutions. As in this case, some may be obtained as limits where some parameters go to $infty$. There are also cases where some particular solutions (singular solutions) are obtained as envelopes of the general solution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does the general solution always give the "main" families of solutions? For example, having graphed a general solution on Desmos with a slider representing each integration constant, letting one slider play at a time produces a smoothly transitioning animated sequence of curves. Is this reliable as a means of seeing all particular solutions except outliers, or can there be entire animated sequences not captured by the general solution?
    $endgroup$
    – user10478
    Dec 31 '18 at 16:11
















5












$begingroup$

This is a very common situation. A general solution for a nonlinear $n$-th order differential equation is a solution depending on $n$ independent parameters, but it does not necessarily give you all the solutions. As in this case, some may be obtained as limits where some parameters go to $infty$. There are also cases where some particular solutions (singular solutions) are obtained as envelopes of the general solution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does the general solution always give the "main" families of solutions? For example, having graphed a general solution on Desmos with a slider representing each integration constant, letting one slider play at a time produces a smoothly transitioning animated sequence of curves. Is this reliable as a means of seeing all particular solutions except outliers, or can there be entire animated sequences not captured by the general solution?
    $endgroup$
    – user10478
    Dec 31 '18 at 16:11














5












5








5





$begingroup$

This is a very common situation. A general solution for a nonlinear $n$-th order differential equation is a solution depending on $n$ independent parameters, but it does not necessarily give you all the solutions. As in this case, some may be obtained as limits where some parameters go to $infty$. There are also cases where some particular solutions (singular solutions) are obtained as envelopes of the general solution.






share|cite|improve this answer









$endgroup$



This is a very common situation. A general solution for a nonlinear $n$-th order differential equation is a solution depending on $n$ independent parameters, but it does not necessarily give you all the solutions. As in this case, some may be obtained as limits where some parameters go to $infty$. There are also cases where some particular solutions (singular solutions) are obtained as envelopes of the general solution.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 31 '18 at 5:27









Robert IsraelRobert Israel

320k23209459




320k23209459












  • $begingroup$
    Does the general solution always give the "main" families of solutions? For example, having graphed a general solution on Desmos with a slider representing each integration constant, letting one slider play at a time produces a smoothly transitioning animated sequence of curves. Is this reliable as a means of seeing all particular solutions except outliers, or can there be entire animated sequences not captured by the general solution?
    $endgroup$
    – user10478
    Dec 31 '18 at 16:11


















  • $begingroup$
    Does the general solution always give the "main" families of solutions? For example, having graphed a general solution on Desmos with a slider representing each integration constant, letting one slider play at a time produces a smoothly transitioning animated sequence of curves. Is this reliable as a means of seeing all particular solutions except outliers, or can there be entire animated sequences not captured by the general solution?
    $endgroup$
    – user10478
    Dec 31 '18 at 16:11
















$begingroup$
Does the general solution always give the "main" families of solutions? For example, having graphed a general solution on Desmos with a slider representing each integration constant, letting one slider play at a time produces a smoothly transitioning animated sequence of curves. Is this reliable as a means of seeing all particular solutions except outliers, or can there be entire animated sequences not captured by the general solution?
$endgroup$
– user10478
Dec 31 '18 at 16:11




$begingroup$
Does the general solution always give the "main" families of solutions? For example, having graphed a general solution on Desmos with a slider representing each integration constant, letting one slider play at a time produces a smoothly transitioning animated sequence of curves. Is this reliable as a means of seeing all particular solutions except outliers, or can there be entire animated sequences not captured by the general solution?
$endgroup$
– user10478
Dec 31 '18 at 16:11











1












$begingroup$

For non-linear equations like this, it is valid to take $Ctoinfty$. You can also write $c = frac{1}{C}$ to get an alternate form



$$ y(t) = frac{1}{t} + frac{2ct}{1-ct^2} $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would we actually have to think of $C$ as some sort of $aleph$, an actual infinite quantity larger than any real number, rather than a finite real approaching $infty$? If we use the usual calculus notion of approaching infinity, it seems we could always find a $|t|$ large enough relative to $C$ that the term doesn't disappear.
    $endgroup$
    – user10478
    Dec 31 '18 at 16:20


















1












$begingroup$

For non-linear equations like this, it is valid to take $Ctoinfty$. You can also write $c = frac{1}{C}$ to get an alternate form



$$ y(t) = frac{1}{t} + frac{2ct}{1-ct^2} $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would we actually have to think of $C$ as some sort of $aleph$, an actual infinite quantity larger than any real number, rather than a finite real approaching $infty$? If we use the usual calculus notion of approaching infinity, it seems we could always find a $|t|$ large enough relative to $C$ that the term doesn't disappear.
    $endgroup$
    – user10478
    Dec 31 '18 at 16:20
















1












1








1





$begingroup$

For non-linear equations like this, it is valid to take $Ctoinfty$. You can also write $c = frac{1}{C}$ to get an alternate form



$$ y(t) = frac{1}{t} + frac{2ct}{1-ct^2} $$






share|cite|improve this answer









$endgroup$



For non-linear equations like this, it is valid to take $Ctoinfty$. You can also write $c = frac{1}{C}$ to get an alternate form



$$ y(t) = frac{1}{t} + frac{2ct}{1-ct^2} $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 31 '18 at 10:40









DylanDylan

12.5k31026




12.5k31026












  • $begingroup$
    Would we actually have to think of $C$ as some sort of $aleph$, an actual infinite quantity larger than any real number, rather than a finite real approaching $infty$? If we use the usual calculus notion of approaching infinity, it seems we could always find a $|t|$ large enough relative to $C$ that the term doesn't disappear.
    $endgroup$
    – user10478
    Dec 31 '18 at 16:20




















  • $begingroup$
    Would we actually have to think of $C$ as some sort of $aleph$, an actual infinite quantity larger than any real number, rather than a finite real approaching $infty$? If we use the usual calculus notion of approaching infinity, it seems we could always find a $|t|$ large enough relative to $C$ that the term doesn't disappear.
    $endgroup$
    – user10478
    Dec 31 '18 at 16:20


















$begingroup$
Would we actually have to think of $C$ as some sort of $aleph$, an actual infinite quantity larger than any real number, rather than a finite real approaching $infty$? If we use the usual calculus notion of approaching infinity, it seems we could always find a $|t|$ large enough relative to $C$ that the term doesn't disappear.
$endgroup$
– user10478
Dec 31 '18 at 16:20






$begingroup$
Would we actually have to think of $C$ as some sort of $aleph$, an actual infinite quantity larger than any real number, rather than a finite real approaching $infty$? If we use the usual calculus notion of approaching infinity, it seems we could always find a $|t|$ large enough relative to $C$ that the term doesn't disappear.
$endgroup$
– user10478
Dec 31 '18 at 16:20




















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