How to solve the following differential equation $B'(t)=1100exp({frac{ln{0.5}}{5}t})+frac{ln{0.5}}{6}B(t)$?












0












$begingroup$


I have constructed this differential equation $B'(t)=1100expleft(frac{ln{0.5}}{5}tright)+frac{ln{0.5}}{6}B(t)$



I have constructed this equation when i asked this question here,
but i'm unable to solve this equation from methods that i know.



I'm unable to separate it, and it's not of the form $y'+p(t)y=q(t)$.



What should be done here?






ordinary-differential-equations derivatives







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    why did you construct this differential equation? (Hint: add context, like your motivation for asking the question, and why this particular question? Why is knowing how to proceed important to you wrt this particular DE?)
    $endgroup$
    – amWhy
    Dec 17 '18 at 19:04








  • 4




    $begingroup$
    "it's not of the form $y'+p(t)y=q(t)$." Is it not? "What should be done here?" Look more carefully at the equation, perhaps.
    $endgroup$
    – Did
    Dec 17 '18 at 19:09






  • 2




    $begingroup$
    $q(t) $ in this case would be $1100expleft(frac{ln(0.5)}{5} tright)$
    $endgroup$
    – amWhy
    Dec 17 '18 at 19:13






  • 1




    $begingroup$
    Yes I've got it.
    $endgroup$
    – user3133165
    Dec 17 '18 at 19:14






  • 2




    $begingroup$
    $ln(0.5)/6$ is a constant function of $t$
    $endgroup$
    – Martín Vacas Vignolo
    Dec 17 '18 at 19:18
















0












$begingroup$


I have constructed this differential equation $B'(t)=1100expleft(frac{ln{0.5}}{5}tright)+frac{ln{0.5}}{6}B(t)$



I have constructed this equation when i asked this question here,
but i'm unable to solve this equation from methods that i know.



I'm unable to separate it, and it's not of the form $y'+p(t)y=q(t)$.



What should be done here?






ordinary-differential-equations derivatives







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    why did you construct this differential equation? (Hint: add context, like your motivation for asking the question, and why this particular question? Why is knowing how to proceed important to you wrt this particular DE?)
    $endgroup$
    – amWhy
    Dec 17 '18 at 19:04








  • 4




    $begingroup$
    "it's not of the form $y'+p(t)y=q(t)$." Is it not? "What should be done here?" Look more carefully at the equation, perhaps.
    $endgroup$
    – Did
    Dec 17 '18 at 19:09






  • 2




    $begingroup$
    $q(t) $ in this case would be $1100expleft(frac{ln(0.5)}{5} tright)$
    $endgroup$
    – amWhy
    Dec 17 '18 at 19:13






  • 1




    $begingroup$
    Yes I've got it.
    $endgroup$
    – user3133165
    Dec 17 '18 at 19:14






  • 2




    $begingroup$
    $ln(0.5)/6$ is a constant function of $t$
    $endgroup$
    – Martín Vacas Vignolo
    Dec 17 '18 at 19:18














0












0








0





$begingroup$


I have constructed this differential equation $B'(t)=1100expleft(frac{ln{0.5}}{5}tright)+frac{ln{0.5}}{6}B(t)$



I have constructed this equation when i asked this question here,
but i'm unable to solve this equation from methods that i know.



I'm unable to separate it, and it's not of the form $y'+p(t)y=q(t)$.



What should be done here?






ordinary-differential-equations derivatives







share|cite|improve this question











$endgroup$




I have constructed this differential equation $B'(t)=1100expleft(frac{ln{0.5}}{5}tright)+frac{ln{0.5}}{6}B(t)$



I have constructed this equation when i asked this question here,
but i'm unable to solve this equation from methods that i know.



I'm unable to separate it, and it's not of the form $y'+p(t)y=q(t)$.



What should be done here?






ordinary-differential-equations derivatives




ordinary-differential-equations derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 19:15









amWhy

192k28225439




192k28225439










asked Dec 17 '18 at 19:02









user3133165user3133165

1838




1838








  • 2




    $begingroup$
    why did you construct this differential equation? (Hint: add context, like your motivation for asking the question, and why this particular question? Why is knowing how to proceed important to you wrt this particular DE?)
    $endgroup$
    – amWhy
    Dec 17 '18 at 19:04








  • 4




    $begingroup$
    "it's not of the form $y'+p(t)y=q(t)$." Is it not? "What should be done here?" Look more carefully at the equation, perhaps.
    $endgroup$
    – Did
    Dec 17 '18 at 19:09






  • 2




    $begingroup$
    $q(t) $ in this case would be $1100expleft(frac{ln(0.5)}{5} tright)$
    $endgroup$
    – amWhy
    Dec 17 '18 at 19:13






  • 1




    $begingroup$
    Yes I've got it.
    $endgroup$
    – user3133165
    Dec 17 '18 at 19:14






  • 2




    $begingroup$
    $ln(0.5)/6$ is a constant function of $t$
    $endgroup$
    – Martín Vacas Vignolo
    Dec 17 '18 at 19:18














  • 2




    $begingroup$
    why did you construct this differential equation? (Hint: add context, like your motivation for asking the question, and why this particular question? Why is knowing how to proceed important to you wrt this particular DE?)
    $endgroup$
    – amWhy
    Dec 17 '18 at 19:04








  • 4




    $begingroup$
    "it's not of the form $y'+p(t)y=q(t)$." Is it not? "What should be done here?" Look more carefully at the equation, perhaps.
    $endgroup$
    – Did
    Dec 17 '18 at 19:09






  • 2




    $begingroup$
    $q(t) $ in this case would be $1100expleft(frac{ln(0.5)}{5} tright)$
    $endgroup$
    – amWhy
    Dec 17 '18 at 19:13






  • 1




    $begingroup$
    Yes I've got it.
    $endgroup$
    – user3133165
    Dec 17 '18 at 19:14






  • 2




    $begingroup$
    $ln(0.5)/6$ is a constant function of $t$
    $endgroup$
    – Martín Vacas Vignolo
    Dec 17 '18 at 19:18








2




2




$begingroup$
why did you construct this differential equation? (Hint: add context, like your motivation for asking the question, and why this particular question? Why is knowing how to proceed important to you wrt this particular DE?)
$endgroup$
– amWhy
Dec 17 '18 at 19:04






$begingroup$
why did you construct this differential equation? (Hint: add context, like your motivation for asking the question, and why this particular question? Why is knowing how to proceed important to you wrt this particular DE?)
$endgroup$
– amWhy
Dec 17 '18 at 19:04






4




4




$begingroup$
"it's not of the form $y'+p(t)y=q(t)$." Is it not? "What should be done here?" Look more carefully at the equation, perhaps.
$endgroup$
– Did
Dec 17 '18 at 19:09




$begingroup$
"it's not of the form $y'+p(t)y=q(t)$." Is it not? "What should be done here?" Look more carefully at the equation, perhaps.
$endgroup$
– Did
Dec 17 '18 at 19:09




2




2




$begingroup$
$q(t) $ in this case would be $1100expleft(frac{ln(0.5)}{5} tright)$
$endgroup$
– amWhy
Dec 17 '18 at 19:13




$begingroup$
$q(t) $ in this case would be $1100expleft(frac{ln(0.5)}{5} tright)$
$endgroup$
– amWhy
Dec 17 '18 at 19:13




1




1




$begingroup$
Yes I've got it.
$endgroup$
– user3133165
Dec 17 '18 at 19:14




$begingroup$
Yes I've got it.
$endgroup$
– user3133165
Dec 17 '18 at 19:14




2




2




$begingroup$
$ln(0.5)/6$ is a constant function of $t$
$endgroup$
– Martín Vacas Vignolo
Dec 17 '18 at 19:18




$begingroup$
$ln(0.5)/6$ is a constant function of $t$
$endgroup$
– Martín Vacas Vignolo
Dec 17 '18 at 19:18










1 Answer
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$begingroup$

The equation is of type $y'+p(t)y=q(t)$.



The solution for $B(t)$ will be $B(t)=frac{int{mu(t)q(t)dt +c}}{mu(t)}$,$mu(t)=e^{int{}p(t)dt}$.






share|cite|improve this answer









$endgroup$













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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    -1












    $begingroup$

    The equation is of type $y'+p(t)y=q(t)$.



    The solution for $B(t)$ will be $B(t)=frac{int{mu(t)q(t)dt +c}}{mu(t)}$,$mu(t)=e^{int{}p(t)dt}$.






    share|cite|improve this answer









    $endgroup$


















      -1












      $begingroup$

      The equation is of type $y'+p(t)y=q(t)$.



      The solution for $B(t)$ will be $B(t)=frac{int{mu(t)q(t)dt +c}}{mu(t)}$,$mu(t)=e^{int{}p(t)dt}$.






      share|cite|improve this answer









      $endgroup$
















        -1












        -1








        -1





        $begingroup$

        The equation is of type $y'+p(t)y=q(t)$.



        The solution for $B(t)$ will be $B(t)=frac{int{mu(t)q(t)dt +c}}{mu(t)}$,$mu(t)=e^{int{}p(t)dt}$.






        share|cite|improve this answer









        $endgroup$



        The equation is of type $y'+p(t)y=q(t)$.



        The solution for $B(t)$ will be $B(t)=frac{int{mu(t)q(t)dt +c}}{mu(t)}$,$mu(t)=e^{int{}p(t)dt}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 9:45









        user3133165user3133165

        1838




        1838






























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