$dy/dx = y sin x-2sin x$, $y(0) = 0$ — Initial Value Problem












0














$$frac{dy}{dx} = ysin x-2sin x, quad y(0) = 0.$$



Initial Value Problem



Hint says: Find an integrating factor






differential-equations







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  • 2




    Welcome to this Math Stack Exchange. Notice that your "question" is kind of a rude implication that the community is just expected to compute your answer, like a robot. It's more helpful if you actually ask a question, and give as much information as you can about why you are stuck. For example, do you not know what an "integrating factor " is?
    – alex.jordan
    Oct 4 '12 at 21:05










  • No I do not. What is an integrating factor?
    – Ryan
    Oct 4 '12 at 21:06










  • I still cannot figure this out.
    – Ryan
    Oct 4 '12 at 21:30
















0














$$frac{dy}{dx} = ysin x-2sin x, quad y(0) = 0.$$



Initial Value Problem



Hint says: Find an integrating factor






differential-equations







share|cite|improve this question




















  • 2




    Welcome to this Math Stack Exchange. Notice that your "question" is kind of a rude implication that the community is just expected to compute your answer, like a robot. It's more helpful if you actually ask a question, and give as much information as you can about why you are stuck. For example, do you not know what an "integrating factor " is?
    – alex.jordan
    Oct 4 '12 at 21:05










  • No I do not. What is an integrating factor?
    – Ryan
    Oct 4 '12 at 21:06










  • I still cannot figure this out.
    – Ryan
    Oct 4 '12 at 21:30














0












0








0







$$frac{dy}{dx} = ysin x-2sin x, quad y(0) = 0.$$



Initial Value Problem



Hint says: Find an integrating factor






differential-equations







share|cite|improve this question















$$frac{dy}{dx} = ysin x-2sin x, quad y(0) = 0.$$



Initial Value Problem



Hint says: Find an integrating factor






differential-equations




differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 4 '12 at 21:00









Henry T. Horton

15k54464




15k54464










asked Oct 4 '12 at 20:57









RyanRyan

39148




39148








  • 2




    Welcome to this Math Stack Exchange. Notice that your "question" is kind of a rude implication that the community is just expected to compute your answer, like a robot. It's more helpful if you actually ask a question, and give as much information as you can about why you are stuck. For example, do you not know what an "integrating factor " is?
    – alex.jordan
    Oct 4 '12 at 21:05










  • No I do not. What is an integrating factor?
    – Ryan
    Oct 4 '12 at 21:06










  • I still cannot figure this out.
    – Ryan
    Oct 4 '12 at 21:30














  • 2




    Welcome to this Math Stack Exchange. Notice that your "question" is kind of a rude implication that the community is just expected to compute your answer, like a robot. It's more helpful if you actually ask a question, and give as much information as you can about why you are stuck. For example, do you not know what an "integrating factor " is?
    – alex.jordan
    Oct 4 '12 at 21:05










  • No I do not. What is an integrating factor?
    – Ryan
    Oct 4 '12 at 21:06










  • I still cannot figure this out.
    – Ryan
    Oct 4 '12 at 21:30








2




2




Welcome to this Math Stack Exchange. Notice that your "question" is kind of a rude implication that the community is just expected to compute your answer, like a robot. It's more helpful if you actually ask a question, and give as much information as you can about why you are stuck. For example, do you not know what an "integrating factor " is?
– alex.jordan
Oct 4 '12 at 21:05




Welcome to this Math Stack Exchange. Notice that your "question" is kind of a rude implication that the community is just expected to compute your answer, like a robot. It's more helpful if you actually ask a question, and give as much information as you can about why you are stuck. For example, do you not know what an "integrating factor " is?
– alex.jordan
Oct 4 '12 at 21:05












No I do not. What is an integrating factor?
– Ryan
Oct 4 '12 at 21:06




No I do not. What is an integrating factor?
– Ryan
Oct 4 '12 at 21:06












I still cannot figure this out.
– Ryan
Oct 4 '12 at 21:30




I still cannot figure this out.
– Ryan
Oct 4 '12 at 21:30










4 Answers
4






active

oldest

votes


















2














When we have a linear differential equation $y'+P(x)y=Q(x)$ of order 1 then $mu (x)=e^{int P(x) dx}$will be an integrating factor. In your case, it is $mu(x)=e^{cos(x)}$. Now multiply it both sides of the equation and you see $$d(e^{cos(x)}y)=-2sin(x)e^{cos(x)}$$ The rest is easy.






share|cite|improve this answer





















  • What do I do next?
    – Ryan
    Oct 4 '12 at 21:15










  • @Ryan: Did you get the right answer? :-)
    – mrs
    Oct 5 '12 at 7:17










  • I think with your help, Ryan got the right answer! +1
    – amWhy
    Mar 23 '13 at 0:44



















2














The simplest type of this ODE is that separable rather than linear. So it is not necessary to treat it as a linear ODE to solve.



$dfrac{dy}{dx}=ysin x-2sin x$



$dfrac{dy}{dx}=(y-2)sin x$



$dfrac{dy}{y-2}=sin x~dx$



$intdfrac{dy}{y-2}=intsin x~dx$



$ln(y-2)=-cos x+c$



$y-2=Ce^{-cos x}$



$y=Ce^{-cos x}+2$



$y(0)=0$ :



$Ce^{-1}+2=0$



$C=-2e$



$therefore y=-2ee^{-cos x}+2=2-2e^{1-cos x}$






share|cite|improve this answer





























    2














    If a differential equation has the form



    $$ y'(x)+p(x)y(x)=q(x),, quad (1), $$ then the integrating factor is given by



    $$ m(x)= {rm e}^{int p(x) dx},. $$



    You need to write your ode in the form of $(1)$ and then find the integrating factor. Then just multiply the original equation by the interating factor and integrate



    $$ (m(x)y)'= q(x) Rightarrow frac{d}{dx}(e^{cos(x)}y)=-2sin(x)e^{cos(x)} $$



    $$ Rightarrow e^{cos(x)}y(x)=2int e^{cos(x)}(-sin(x))dx + C = 2e^{cos(x)} +C $$



    $$ y(x)= 2 + C ,{rm e}^{-cos(x)} ,.$$



    To find $C$, you need to use the initial condition $y(0)=0$,



    $$ y(0) = 0 = 2 + C{rm e}^{-cos(0)} Rightarrow C = -2 {rm e}$$



    Substituting the value of $C$ in the solution gives



    $$ y(x)= 2 - 2 ,{rm e}^{1-cos(x)} ,.$$






    share|cite|improve this answer























    • So I got the integral, but how would I integrate: e^(coax)y(x)?
      – Ryan
      Oct 4 '12 at 21:20



















    0














    Hint: Write it as $frac{y'}{y-2}=sin(x)$ and integrate.






    share|cite|improve this answer





















    • The answer above is correct. However, if the person asking the question insists you use the hint, then you write the ODE as $
      – Stefan Smith
      Oct 4 '12 at 21:06











    Your Answer





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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    When we have a linear differential equation $y'+P(x)y=Q(x)$ of order 1 then $mu (x)=e^{int P(x) dx}$will be an integrating factor. In your case, it is $mu(x)=e^{cos(x)}$. Now multiply it both sides of the equation and you see $$d(e^{cos(x)}y)=-2sin(x)e^{cos(x)}$$ The rest is easy.






    share|cite|improve this answer





















    • What do I do next?
      – Ryan
      Oct 4 '12 at 21:15










    • @Ryan: Did you get the right answer? :-)
      – mrs
      Oct 5 '12 at 7:17










    • I think with your help, Ryan got the right answer! +1
      – amWhy
      Mar 23 '13 at 0:44
















    2














    When we have a linear differential equation $y'+P(x)y=Q(x)$ of order 1 then $mu (x)=e^{int P(x) dx}$will be an integrating factor. In your case, it is $mu(x)=e^{cos(x)}$. Now multiply it both sides of the equation and you see $$d(e^{cos(x)}y)=-2sin(x)e^{cos(x)}$$ The rest is easy.






    share|cite|improve this answer





















    • What do I do next?
      – Ryan
      Oct 4 '12 at 21:15










    • @Ryan: Did you get the right answer? :-)
      – mrs
      Oct 5 '12 at 7:17










    • I think with your help, Ryan got the right answer! +1
      – amWhy
      Mar 23 '13 at 0:44














    2












    2








    2






    When we have a linear differential equation $y'+P(x)y=Q(x)$ of order 1 then $mu (x)=e^{int P(x) dx}$will be an integrating factor. In your case, it is $mu(x)=e^{cos(x)}$. Now multiply it both sides of the equation and you see $$d(e^{cos(x)}y)=-2sin(x)e^{cos(x)}$$ The rest is easy.






    share|cite|improve this answer












    When we have a linear differential equation $y'+P(x)y=Q(x)$ of order 1 then $mu (x)=e^{int P(x) dx}$will be an integrating factor. In your case, it is $mu(x)=e^{cos(x)}$. Now multiply it both sides of the equation and you see $$d(e^{cos(x)}y)=-2sin(x)e^{cos(x)}$$ The rest is easy.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 4 '12 at 21:08









    mrsmrs

    1




    1












    • What do I do next?
      – Ryan
      Oct 4 '12 at 21:15










    • @Ryan: Did you get the right answer? :-)
      – mrs
      Oct 5 '12 at 7:17










    • I think with your help, Ryan got the right answer! +1
      – amWhy
      Mar 23 '13 at 0:44


















    • What do I do next?
      – Ryan
      Oct 4 '12 at 21:15










    • @Ryan: Did you get the right answer? :-)
      – mrs
      Oct 5 '12 at 7:17










    • I think with your help, Ryan got the right answer! +1
      – amWhy
      Mar 23 '13 at 0:44
















    What do I do next?
    – Ryan
    Oct 4 '12 at 21:15




    What do I do next?
    – Ryan
    Oct 4 '12 at 21:15












    @Ryan: Did you get the right answer? :-)
    – mrs
    Oct 5 '12 at 7:17




    @Ryan: Did you get the right answer? :-)
    – mrs
    Oct 5 '12 at 7:17












    I think with your help, Ryan got the right answer! +1
    – amWhy
    Mar 23 '13 at 0:44




    I think with your help, Ryan got the right answer! +1
    – amWhy
    Mar 23 '13 at 0:44











    2














    The simplest type of this ODE is that separable rather than linear. So it is not necessary to treat it as a linear ODE to solve.



    $dfrac{dy}{dx}=ysin x-2sin x$



    $dfrac{dy}{dx}=(y-2)sin x$



    $dfrac{dy}{y-2}=sin x~dx$



    $intdfrac{dy}{y-2}=intsin x~dx$



    $ln(y-2)=-cos x+c$



    $y-2=Ce^{-cos x}$



    $y=Ce^{-cos x}+2$



    $y(0)=0$ :



    $Ce^{-1}+2=0$



    $C=-2e$



    $therefore y=-2ee^{-cos x}+2=2-2e^{1-cos x}$






    share|cite|improve this answer


























      2














      The simplest type of this ODE is that separable rather than linear. So it is not necessary to treat it as a linear ODE to solve.



      $dfrac{dy}{dx}=ysin x-2sin x$



      $dfrac{dy}{dx}=(y-2)sin x$



      $dfrac{dy}{y-2}=sin x~dx$



      $intdfrac{dy}{y-2}=intsin x~dx$



      $ln(y-2)=-cos x+c$



      $y-2=Ce^{-cos x}$



      $y=Ce^{-cos x}+2$



      $y(0)=0$ :



      $Ce^{-1}+2=0$



      $C=-2e$



      $therefore y=-2ee^{-cos x}+2=2-2e^{1-cos x}$






      share|cite|improve this answer
























        2












        2








        2






        The simplest type of this ODE is that separable rather than linear. So it is not necessary to treat it as a linear ODE to solve.



        $dfrac{dy}{dx}=ysin x-2sin x$



        $dfrac{dy}{dx}=(y-2)sin x$



        $dfrac{dy}{y-2}=sin x~dx$



        $intdfrac{dy}{y-2}=intsin x~dx$



        $ln(y-2)=-cos x+c$



        $y-2=Ce^{-cos x}$



        $y=Ce^{-cos x}+2$



        $y(0)=0$ :



        $Ce^{-1}+2=0$



        $C=-2e$



        $therefore y=-2ee^{-cos x}+2=2-2e^{1-cos x}$






        share|cite|improve this answer












        The simplest type of this ODE is that separable rather than linear. So it is not necessary to treat it as a linear ODE to solve.



        $dfrac{dy}{dx}=ysin x-2sin x$



        $dfrac{dy}{dx}=(y-2)sin x$



        $dfrac{dy}{y-2}=sin x~dx$



        $intdfrac{dy}{y-2}=intsin x~dx$



        $ln(y-2)=-cos x+c$



        $y-2=Ce^{-cos x}$



        $y=Ce^{-cos x}+2$



        $y(0)=0$ :



        $Ce^{-1}+2=0$



        $C=-2e$



        $therefore y=-2ee^{-cos x}+2=2-2e^{1-cos x}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 4 '12 at 23:37









        doraemonpauldoraemonpaul

        12.5k31660




        12.5k31660























            2














            If a differential equation has the form



            $$ y'(x)+p(x)y(x)=q(x),, quad (1), $$ then the integrating factor is given by



            $$ m(x)= {rm e}^{int p(x) dx},. $$



            You need to write your ode in the form of $(1)$ and then find the integrating factor. Then just multiply the original equation by the interating factor and integrate



            $$ (m(x)y)'= q(x) Rightarrow frac{d}{dx}(e^{cos(x)}y)=-2sin(x)e^{cos(x)} $$



            $$ Rightarrow e^{cos(x)}y(x)=2int e^{cos(x)}(-sin(x))dx + C = 2e^{cos(x)} +C $$



            $$ y(x)= 2 + C ,{rm e}^{-cos(x)} ,.$$



            To find $C$, you need to use the initial condition $y(0)=0$,



            $$ y(0) = 0 = 2 + C{rm e}^{-cos(0)} Rightarrow C = -2 {rm e}$$



            Substituting the value of $C$ in the solution gives



            $$ y(x)= 2 - 2 ,{rm e}^{1-cos(x)} ,.$$






            share|cite|improve this answer























            • So I got the integral, but how would I integrate: e^(coax)y(x)?
              – Ryan
              Oct 4 '12 at 21:20
















            2














            If a differential equation has the form



            $$ y'(x)+p(x)y(x)=q(x),, quad (1), $$ then the integrating factor is given by



            $$ m(x)= {rm e}^{int p(x) dx},. $$



            You need to write your ode in the form of $(1)$ and then find the integrating factor. Then just multiply the original equation by the interating factor and integrate



            $$ (m(x)y)'= q(x) Rightarrow frac{d}{dx}(e^{cos(x)}y)=-2sin(x)e^{cos(x)} $$



            $$ Rightarrow e^{cos(x)}y(x)=2int e^{cos(x)}(-sin(x))dx + C = 2e^{cos(x)} +C $$



            $$ y(x)= 2 + C ,{rm e}^{-cos(x)} ,.$$



            To find $C$, you need to use the initial condition $y(0)=0$,



            $$ y(0) = 0 = 2 + C{rm e}^{-cos(0)} Rightarrow C = -2 {rm e}$$



            Substituting the value of $C$ in the solution gives



            $$ y(x)= 2 - 2 ,{rm e}^{1-cos(x)} ,.$$






            share|cite|improve this answer























            • So I got the integral, but how would I integrate: e^(coax)y(x)?
              – Ryan
              Oct 4 '12 at 21:20














            2












            2








            2






            If a differential equation has the form



            $$ y'(x)+p(x)y(x)=q(x),, quad (1), $$ then the integrating factor is given by



            $$ m(x)= {rm e}^{int p(x) dx},. $$



            You need to write your ode in the form of $(1)$ and then find the integrating factor. Then just multiply the original equation by the interating factor and integrate



            $$ (m(x)y)'= q(x) Rightarrow frac{d}{dx}(e^{cos(x)}y)=-2sin(x)e^{cos(x)} $$



            $$ Rightarrow e^{cos(x)}y(x)=2int e^{cos(x)}(-sin(x))dx + C = 2e^{cos(x)} +C $$



            $$ y(x)= 2 + C ,{rm e}^{-cos(x)} ,.$$



            To find $C$, you need to use the initial condition $y(0)=0$,



            $$ y(0) = 0 = 2 + C{rm e}^{-cos(0)} Rightarrow C = -2 {rm e}$$



            Substituting the value of $C$ in the solution gives



            $$ y(x)= 2 - 2 ,{rm e}^{1-cos(x)} ,.$$






            share|cite|improve this answer














            If a differential equation has the form



            $$ y'(x)+p(x)y(x)=q(x),, quad (1), $$ then the integrating factor is given by



            $$ m(x)= {rm e}^{int p(x) dx},. $$



            You need to write your ode in the form of $(1)$ and then find the integrating factor. Then just multiply the original equation by the interating factor and integrate



            $$ (m(x)y)'= q(x) Rightarrow frac{d}{dx}(e^{cos(x)}y)=-2sin(x)e^{cos(x)} $$



            $$ Rightarrow e^{cos(x)}y(x)=2int e^{cos(x)}(-sin(x))dx + C = 2e^{cos(x)} +C $$



            $$ y(x)= 2 + C ,{rm e}^{-cos(x)} ,.$$



            To find $C$, you need to use the initial condition $y(0)=0$,



            $$ y(0) = 0 = 2 + C{rm e}^{-cos(0)} Rightarrow C = -2 {rm e}$$



            Substituting the value of $C$ in the solution gives



            $$ y(x)= 2 - 2 ,{rm e}^{1-cos(x)} ,.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Oct 6 '12 at 5:26

























            answered Oct 4 '12 at 21:16









            Mhenni BenghorbalMhenni Benghorbal

            43.1k63574




            43.1k63574












            • So I got the integral, but how would I integrate: e^(coax)y(x)?
              – Ryan
              Oct 4 '12 at 21:20


















            • So I got the integral, but how would I integrate: e^(coax)y(x)?
              – Ryan
              Oct 4 '12 at 21:20
















            So I got the integral, but how would I integrate: e^(coax)y(x)?
            – Ryan
            Oct 4 '12 at 21:20




            So I got the integral, but how would I integrate: e^(coax)y(x)?
            – Ryan
            Oct 4 '12 at 21:20











            0














            Hint: Write it as $frac{y'}{y-2}=sin(x)$ and integrate.






            share|cite|improve this answer





















            • The answer above is correct. However, if the person asking the question insists you use the hint, then you write the ODE as $
              – Stefan Smith
              Oct 4 '12 at 21:06
















            0














            Hint: Write it as $frac{y'}{y-2}=sin(x)$ and integrate.






            share|cite|improve this answer





















            • The answer above is correct. However, if the person asking the question insists you use the hint, then you write the ODE as $
              – Stefan Smith
              Oct 4 '12 at 21:06














            0












            0








            0






            Hint: Write it as $frac{y'}{y-2}=sin(x)$ and integrate.






            share|cite|improve this answer












            Hint: Write it as $frac{y'}{y-2}=sin(x)$ and integrate.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 4 '12 at 21:02









            draks ...draks ...

            11.5k644128




            11.5k644128












            • The answer above is correct. However, if the person asking the question insists you use the hint, then you write the ODE as $
              – Stefan Smith
              Oct 4 '12 at 21:06


















            • The answer above is correct. However, if the person asking the question insists you use the hint, then you write the ODE as $
              – Stefan Smith
              Oct 4 '12 at 21:06
















            The answer above is correct. However, if the person asking the question insists you use the hint, then you write the ODE as $
            – Stefan Smith
            Oct 4 '12 at 21:06




            The answer above is correct. However, if the person asking the question insists you use the hint, then you write the ODE as $
            – Stefan Smith
            Oct 4 '12 at 21:06


















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            Grease: Live!

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            A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
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            When does type information flow backwards in C++?

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            up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
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