Find the particular solution of, $y=Ce^{-2x}+De^{-3x}+cos(x)+sin(x)$ [on hold]











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Where, $y = 1$, $dy/dx = 0$ when $x = 0$.



I've tried using simultaneous equations but keep getting $0$ as an answer for both constants, not sure how else to proceed.






differential-equations







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put on hold as off-topic by amWhy, max_zorn, José Carlos Santos, Zvi, Christopher yesterday


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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, max_zorn, José Carlos Santos, Christopher

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  • Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
    – RocketKangaroo
    2 days ago










  • I get that $C=-1$ and $D=1$. If both $C=0$ and $D=0$, then $y=cos(x)+sin(x)$, and $y'=-sin(x)+cos(x)$, so $y'(0)=1$, which is a contradiction.
    – projectilemotion
    2 days ago










  • @projectilemotion: yep, sorry, my bad.
    – Yves Daoust
    2 days ago















up vote
0
down vote

favorite












Where, $y = 1$, $dy/dx = 0$ when $x = 0$.



I've tried using simultaneous equations but keep getting $0$ as an answer for both constants, not sure how else to proceed.






differential-equations







share|cite|improve this question









New contributor




RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as off-topic by amWhy, max_zorn, José Carlos Santos, Zvi, Christopher yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, max_zorn, José Carlos Santos, Christopher

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
    – RocketKangaroo
    2 days ago










  • I get that $C=-1$ and $D=1$. If both $C=0$ and $D=0$, then $y=cos(x)+sin(x)$, and $y'=-sin(x)+cos(x)$, so $y'(0)=1$, which is a contradiction.
    – projectilemotion
    2 days ago










  • @projectilemotion: yep, sorry, my bad.
    – Yves Daoust
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Where, $y = 1$, $dy/dx = 0$ when $x = 0$.



I've tried using simultaneous equations but keep getting $0$ as an answer for both constants, not sure how else to proceed.






differential-equations







share|cite|improve this question









New contributor




RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Where, $y = 1$, $dy/dx = 0$ when $x = 0$.



I've tried using simultaneous equations but keep getting $0$ as an answer for both constants, not sure how else to proceed.






differential-equations




differential-equations






share|cite|improve this question









New contributor




RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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edited 2 days ago









mrtaurho

2,3691726




2,3691726






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RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 days ago









RocketKangaroo

163




163




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RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






RocketKangaroo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




put on hold as off-topic by amWhy, max_zorn, José Carlos Santos, Zvi, Christopher yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, max_zorn, José Carlos Santos, Christopher

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by amWhy, max_zorn, José Carlos Santos, Zvi, Christopher yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, max_zorn, José Carlos Santos, Christopher

If this question can be reworded to fit the rules in the help center, please edit the question.












  • Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
    – RocketKangaroo
    2 days ago










  • I get that $C=-1$ and $D=1$. If both $C=0$ and $D=0$, then $y=cos(x)+sin(x)$, and $y'=-sin(x)+cos(x)$, so $y'(0)=1$, which is a contradiction.
    – projectilemotion
    2 days ago










  • @projectilemotion: yep, sorry, my bad.
    – Yves Daoust
    2 days ago


















  • Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
    – RocketKangaroo
    2 days ago










  • I get that $C=-1$ and $D=1$. If both $C=0$ and $D=0$, then $y=cos(x)+sin(x)$, and $y'=-sin(x)+cos(x)$, so $y'(0)=1$, which is a contradiction.
    – projectilemotion
    2 days ago










  • @projectilemotion: yep, sorry, my bad.
    – Yves Daoust
    2 days ago
















Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
– RocketKangaroo
2 days ago




Literally just edited :D, i think my skepticism is rooted in the fact that it appears too good to be true
– RocketKangaroo
2 days ago












I get that $C=-1$ and $D=1$. If both $C=0$ and $D=0$, then $y=cos(x)+sin(x)$, and $y'=-sin(x)+cos(x)$, so $y'(0)=1$, which is a contradiction.
– projectilemotion
2 days ago




I get that $C=-1$ and $D=1$. If both $C=0$ and $D=0$, then $y=cos(x)+sin(x)$, and $y'=-sin(x)+cos(x)$, so $y'(0)=1$, which is a contradiction.
– projectilemotion
2 days ago












@projectilemotion: yep, sorry, my bad.
– Yves Daoust
2 days ago




@projectilemotion: yep, sorry, my bad.
– Yves Daoust
2 days ago










1 Answer
1






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up vote
3
down vote













From the initial conditions,



$$begin{cases}1=C+D+1,
\0=-2C-3D+1.end{cases}$$



$C=D=0$ is obviously not a solution.






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote













    From the initial conditions,



    $$begin{cases}1=C+D+1,
    \0=-2C-3D+1.end{cases}$$



    $C=D=0$ is obviously not a solution.






    share|cite|improve this answer

























      up vote
      3
      down vote













      From the initial conditions,



      $$begin{cases}1=C+D+1,
      \0=-2C-3D+1.end{cases}$$



      $C=D=0$ is obviously not a solution.






      share|cite|improve this answer























        up vote
        3
        down vote










        up vote
        3
        down vote









        From the initial conditions,



        $$begin{cases}1=C+D+1,
        \0=-2C-3D+1.end{cases}$$



        $C=D=0$ is obviously not a solution.






        share|cite|improve this answer












        From the initial conditions,



        $$begin{cases}1=C+D+1,
        \0=-2C-3D+1.end{cases}$$



        $C=D=0$ is obviously not a solution.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Yves Daoust

        121k668216




        121k668216















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