$dy/dx = y sin x-2sin x$, $y(0) = 0$ — 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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            Are they similar matrix

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            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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