Help with integral and constants












0












$begingroup$


I am trying to understand an integral in a problem I am doing:



$intfrac {du}{u} = -intfrac {dt}{t}$



Which according to the textbook solutions gives $ln(u) = -ln(t) + ln(C)$



This seems to be done for convenience, because we get $u(t) = frac{C}{t}$ as the final answer. But when I integrate, I get $- ln(t) -ln(C)$, which gives $u(t) = frac{1}{Ct}$. Is it just the case that we can make the constant of integration whatever we like in problems like this? I think it is strange that the choice of constant would change the final answer.



Thanks for your help.






calculus integration







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$endgroup$








  • 1




    $begingroup$
    The constant of integration is entirely arbitrary. The two results are the same, just with different constants. Write $$u(t)={1over Kt}$$ and you'll see it's the same as the prior result; we just have $K={1over C}$
    $endgroup$
    – saulspatz
    Dec 2 '18 at 15:14










  • $begingroup$
    we are able to redefine the integration constant however we want, for example such that $ln(C_1)=-ln(C_2)$
    $endgroup$
    – Henry Lee
    Dec 2 '18 at 18:32
















0












$begingroup$


I am trying to understand an integral in a problem I am doing:



$intfrac {du}{u} = -intfrac {dt}{t}$



Which according to the textbook solutions gives $ln(u) = -ln(t) + ln(C)$



This seems to be done for convenience, because we get $u(t) = frac{C}{t}$ as the final answer. But when I integrate, I get $- ln(t) -ln(C)$, which gives $u(t) = frac{1}{Ct}$. Is it just the case that we can make the constant of integration whatever we like in problems like this? I think it is strange that the choice of constant would change the final answer.



Thanks for your help.






calculus integration







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The constant of integration is entirely arbitrary. The two results are the same, just with different constants. Write $$u(t)={1over Kt}$$ and you'll see it's the same as the prior result; we just have $K={1over C}$
    $endgroup$
    – saulspatz
    Dec 2 '18 at 15:14










  • $begingroup$
    we are able to redefine the integration constant however we want, for example such that $ln(C_1)=-ln(C_2)$
    $endgroup$
    – Henry Lee
    Dec 2 '18 at 18:32














0












0








0





$begingroup$


I am trying to understand an integral in a problem I am doing:



$intfrac {du}{u} = -intfrac {dt}{t}$



Which according to the textbook solutions gives $ln(u) = -ln(t) + ln(C)$



This seems to be done for convenience, because we get $u(t) = frac{C}{t}$ as the final answer. But when I integrate, I get $- ln(t) -ln(C)$, which gives $u(t) = frac{1}{Ct}$. Is it just the case that we can make the constant of integration whatever we like in problems like this? I think it is strange that the choice of constant would change the final answer.



Thanks for your help.






calculus integration







share|cite|improve this question









$endgroup$




I am trying to understand an integral in a problem I am doing:



$intfrac {du}{u} = -intfrac {dt}{t}$



Which according to the textbook solutions gives $ln(u) = -ln(t) + ln(C)$



This seems to be done for convenience, because we get $u(t) = frac{C}{t}$ as the final answer. But when I integrate, I get $- ln(t) -ln(C)$, which gives $u(t) = frac{1}{Ct}$. Is it just the case that we can make the constant of integration whatever we like in problems like this? I think it is strange that the choice of constant would change the final answer.



Thanks for your help.






calculus integration




calculus integration






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share|cite|improve this question











share|cite|improve this question




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asked Dec 2 '18 at 15:08









ChristianChristian

367




367








  • 1




    $begingroup$
    The constant of integration is entirely arbitrary. The two results are the same, just with different constants. Write $$u(t)={1over Kt}$$ and you'll see it's the same as the prior result; we just have $K={1over C}$
    $endgroup$
    – saulspatz
    Dec 2 '18 at 15:14










  • $begingroup$
    we are able to redefine the integration constant however we want, for example such that $ln(C_1)=-ln(C_2)$
    $endgroup$
    – Henry Lee
    Dec 2 '18 at 18:32














  • 1




    $begingroup$
    The constant of integration is entirely arbitrary. The two results are the same, just with different constants. Write $$u(t)={1over Kt}$$ and you'll see it's the same as the prior result; we just have $K={1over C}$
    $endgroup$
    – saulspatz
    Dec 2 '18 at 15:14










  • $begingroup$
    we are able to redefine the integration constant however we want, for example such that $ln(C_1)=-ln(C_2)$
    $endgroup$
    – Henry Lee
    Dec 2 '18 at 18:32








1




1




$begingroup$
The constant of integration is entirely arbitrary. The two results are the same, just with different constants. Write $$u(t)={1over Kt}$$ and you'll see it's the same as the prior result; we just have $K={1over C}$
$endgroup$
– saulspatz
Dec 2 '18 at 15:14




$begingroup$
The constant of integration is entirely arbitrary. The two results are the same, just with different constants. Write $$u(t)={1over Kt}$$ and you'll see it's the same as the prior result; we just have $K={1over C}$
$endgroup$
– saulspatz
Dec 2 '18 at 15:14












$begingroup$
we are able to redefine the integration constant however we want, for example such that $ln(C_1)=-ln(C_2)$
$endgroup$
– Henry Lee
Dec 2 '18 at 18:32




$begingroup$
we are able to redefine the integration constant however we want, for example such that $ln(C_1)=-ln(C_2)$
$endgroup$
– Henry Lee
Dec 2 '18 at 18:32










1 Answer
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$begingroup$

C is an arbitrary constant, if you call C0 the one in the textbook solution and C1 the one you get when you integrate (so we know which one is which).



C1=1/C0



And then -ln(C1) becomes +ln(C0) and the result u(t)=1/(C1t) becomes u(t)=C0/t






share|cite|improve this answer









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    2












    $begingroup$

    C is an arbitrary constant, if you call C0 the one in the textbook solution and C1 the one you get when you integrate (so we know which one is which).



    C1=1/C0



    And then -ln(C1) becomes +ln(C0) and the result u(t)=1/(C1t) becomes u(t)=C0/t






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      C is an arbitrary constant, if you call C0 the one in the textbook solution and C1 the one you get when you integrate (so we know which one is which).



      C1=1/C0



      And then -ln(C1) becomes +ln(C0) and the result u(t)=1/(C1t) becomes u(t)=C0/t






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        C is an arbitrary constant, if you call C0 the one in the textbook solution and C1 the one you get when you integrate (so we know which one is which).



        C1=1/C0



        And then -ln(C1) becomes +ln(C0) and the result u(t)=1/(C1t) becomes u(t)=C0/t






        share|cite|improve this answer









        $endgroup$



        C is an arbitrary constant, if you call C0 the one in the textbook solution and C1 the one you get when you integrate (so we know which one is which).



        C1=1/C0



        And then -ln(C1) becomes +ln(C0) and the result u(t)=1/(C1t) becomes u(t)=C0/t







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 15:17









        Eponyme WebEponyme Web

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