How To Solve a Trigonometric Differential Equation












2














Salutations, I have been trying to approach an ODE with trigonometric functions that I found interesting:



$$y'+xsin(2y)=xe^{-x^2}cos^2(y)$$



I tried to find a result with wolfram web page (free version) and I got this one:



$$y=arctanleft(frac{1}{2}e^{-x^2}(c+x^2)right)$$



I have tried to approach this exercise by substitution of variables, also separable variables and I have not had luck by power series,
and I do not know if methods like those of Ricatti and Bernoulli are appropriate for this case.



This is just for academic curiosity and I would like to understand better this kind of ODEs. So, I require any guidance or starting steps or explanations about how to approach this kind of exercises.



Thanks for your attention.






differential-equations trigonometry proof-explanation







share|cite|improve this question



























    2














    Salutations, I have been trying to approach an ODE with trigonometric functions that I found interesting:



    $$y'+xsin(2y)=xe^{-x^2}cos^2(y)$$



    I tried to find a result with wolfram web page (free version) and I got this one:



    $$y=arctanleft(frac{1}{2}e^{-x^2}(c+x^2)right)$$



    I have tried to approach this exercise by substitution of variables, also separable variables and I have not had luck by power series,
    and I do not know if methods like those of Ricatti and Bernoulli are appropriate for this case.



    This is just for academic curiosity and I would like to understand better this kind of ODEs. So, I require any guidance or starting steps or explanations about how to approach this kind of exercises.



    Thanks for your attention.






    differential-equations trigonometry proof-explanation







    share|cite|improve this question

























      2












      2








      2


      2





      Salutations, I have been trying to approach an ODE with trigonometric functions that I found interesting:



      $$y'+xsin(2y)=xe^{-x^2}cos^2(y)$$



      I tried to find a result with wolfram web page (free version) and I got this one:



      $$y=arctanleft(frac{1}{2}e^{-x^2}(c+x^2)right)$$



      I have tried to approach this exercise by substitution of variables, also separable variables and I have not had luck by power series,
      and I do not know if methods like those of Ricatti and Bernoulli are appropriate for this case.



      This is just for academic curiosity and I would like to understand better this kind of ODEs. So, I require any guidance or starting steps or explanations about how to approach this kind of exercises.



      Thanks for your attention.






      differential-equations trigonometry proof-explanation







      share|cite|improve this question













      Salutations, I have been trying to approach an ODE with trigonometric functions that I found interesting:



      $$y'+xsin(2y)=xe^{-x^2}cos^2(y)$$



      I tried to find a result with wolfram web page (free version) and I got this one:



      $$y=arctanleft(frac{1}{2}e^{-x^2}(c+x^2)right)$$



      I have tried to approach this exercise by substitution of variables, also separable variables and I have not had luck by power series,
      and I do not know if methods like those of Ricatti and Bernoulli are appropriate for this case.



      This is just for academic curiosity and I would like to understand better this kind of ODEs. So, I require any guidance or starting steps or explanations about how to approach this kind of exercises.



      Thanks for your attention.






      differential-equations trigonometry proof-explanation




      differential-equations trigonometry proof-explanation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      ht1204

      294




      294






















          1 Answer
          1






          active

          oldest

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          7














          This is in general so non-linear that you can not expect a closed solution. However, as it is an exercise a closed solution most probably exists, so you have to consider the parts of this equation. With some experience one may see that dividing by $cos^2y$ gives
          $$
          frac{y'}{cos^2 y}+2xtan y=xe^{-x^2}
          $$
          which has the form
          $$
          f'(y)y'+2xf(y)=xe^{-x^2}
          $$
          which now is linear in $u=f(y)=tan(y)$. For this linear equation, $e^{x^2}$ is an integrating factor which miraculously also simplifies the right side. After integrating you get
          $$
          e^{x^2}tan(y(x))=frac12x^2+c.
          $$






          share|cite|improve this answer





















          • Thanks very much for your commentary, I was taking unnecessary and hard paths to solve this exercise, thanks again for your explanation, the substitution of tangent function was very helpful, thanks for all.
            – ht1204
            54 mins ago











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          7














          This is in general so non-linear that you can not expect a closed solution. However, as it is an exercise a closed solution most probably exists, so you have to consider the parts of this equation. With some experience one may see that dividing by $cos^2y$ gives
          $$
          frac{y'}{cos^2 y}+2xtan y=xe^{-x^2}
          $$
          which has the form
          $$
          f'(y)y'+2xf(y)=xe^{-x^2}
          $$
          which now is linear in $u=f(y)=tan(y)$. For this linear equation, $e^{x^2}$ is an integrating factor which miraculously also simplifies the right side. After integrating you get
          $$
          e^{x^2}tan(y(x))=frac12x^2+c.
          $$






          share|cite|improve this answer





















          • Thanks very much for your commentary, I was taking unnecessary and hard paths to solve this exercise, thanks again for your explanation, the substitution of tangent function was very helpful, thanks for all.
            – ht1204
            54 mins ago
















          7














          This is in general so non-linear that you can not expect a closed solution. However, as it is an exercise a closed solution most probably exists, so you have to consider the parts of this equation. With some experience one may see that dividing by $cos^2y$ gives
          $$
          frac{y'}{cos^2 y}+2xtan y=xe^{-x^2}
          $$
          which has the form
          $$
          f'(y)y'+2xf(y)=xe^{-x^2}
          $$
          which now is linear in $u=f(y)=tan(y)$. For this linear equation, $e^{x^2}$ is an integrating factor which miraculously also simplifies the right side. After integrating you get
          $$
          e^{x^2}tan(y(x))=frac12x^2+c.
          $$






          share|cite|improve this answer





















          • Thanks very much for your commentary, I was taking unnecessary and hard paths to solve this exercise, thanks again for your explanation, the substitution of tangent function was very helpful, thanks for all.
            – ht1204
            54 mins ago














          7












          7








          7






          This is in general so non-linear that you can not expect a closed solution. However, as it is an exercise a closed solution most probably exists, so you have to consider the parts of this equation. With some experience one may see that dividing by $cos^2y$ gives
          $$
          frac{y'}{cos^2 y}+2xtan y=xe^{-x^2}
          $$
          which has the form
          $$
          f'(y)y'+2xf(y)=xe^{-x^2}
          $$
          which now is linear in $u=f(y)=tan(y)$. For this linear equation, $e^{x^2}$ is an integrating factor which miraculously also simplifies the right side. After integrating you get
          $$
          e^{x^2}tan(y(x))=frac12x^2+c.
          $$






          share|cite|improve this answer












          This is in general so non-linear that you can not expect a closed solution. However, as it is an exercise a closed solution most probably exists, so you have to consider the parts of this equation. With some experience one may see that dividing by $cos^2y$ gives
          $$
          frac{y'}{cos^2 y}+2xtan y=xe^{-x^2}
          $$
          which has the form
          $$
          f'(y)y'+2xf(y)=xe^{-x^2}
          $$
          which now is linear in $u=f(y)=tan(y)$. For this linear equation, $e^{x^2}$ is an integrating factor which miraculously also simplifies the right side. After integrating you get
          $$
          e^{x^2}tan(y(x))=frac12x^2+c.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          LutzL

          55.6k42053




          55.6k42053












          • Thanks very much for your commentary, I was taking unnecessary and hard paths to solve this exercise, thanks again for your explanation, the substitution of tangent function was very helpful, thanks for all.
            – ht1204
            54 mins ago


















          • Thanks very much for your commentary, I was taking unnecessary and hard paths to solve this exercise, thanks again for your explanation, the substitution of tangent function was very helpful, thanks for all.
            – ht1204
            54 mins ago
















          Thanks very much for your commentary, I was taking unnecessary and hard paths to solve this exercise, thanks again for your explanation, the substitution of tangent function was very helpful, thanks for all.
          – ht1204
          54 mins ago




          Thanks very much for your commentary, I was taking unnecessary and hard paths to solve this exercise, thanks again for your explanation, the substitution of tangent function was very helpful, thanks for all.
          – ht1204
          54 mins ago


















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