While solving LDE can the IF $e^{int frac{dx}{x}}$ be taken as x instead of |x|?












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While solving the differential equation
$xfrac {dy}{dx}+y-x+ xycot{x}=0;for;xneq 0$, I noticed in the attached solution, the omission of the absolute value symbol around the integration factor $xsin{x}$. To me it appeared as if half the solution set is ignored by the attached solution where $xsin{x}<0$. However I need some experts' opinion and to be sure if my worries are uncalled for.
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    $begingroup$


    While solving the differential equation
    $xfrac {dy}{dx}+y-x+ xycot{x}=0;for;xneq 0$, I noticed in the attached solution, the omission of the absolute value symbol around the integration factor $xsin{x}$. To me it appeared as if half the solution set is ignored by the attached solution where $xsin{x}<0$. However I need some experts' opinion and to be sure if my worries are uncalled for.
    enter image description here



    enter image description here










    share|cite|improve this question









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      1





      $begingroup$


      While solving the differential equation
      $xfrac {dy}{dx}+y-x+ xycot{x}=0;for;xneq 0$, I noticed in the attached solution, the omission of the absolute value symbol around the integration factor $xsin{x}$. To me it appeared as if half the solution set is ignored by the attached solution where $xsin{x}<0$. However I need some experts' opinion and to be sure if my worries are uncalled for.
      enter image description here



      enter image description here










      share|cite|improve this question









      $endgroup$




      While solving the differential equation
      $xfrac {dy}{dx}+y-x+ xycot{x}=0;for;xneq 0$, I noticed in the attached solution, the omission of the absolute value symbol around the integration factor $xsin{x}$. To me it appeared as if half the solution set is ignored by the attached solution where $xsin{x}<0$. However I need some experts' opinion and to be sure if my worries are uncalled for.
      enter image description here



      enter image description here







      calculus ordinary-differential-equations






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      asked Jan 18 at 17:14









      BibThePhysicistBibThePhysicist

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          No solution is lost here. It suffices to find ONE integrating factor. So $xsin(x)$ is enough. Nothing changes if we take $-xsin(x)$ or $Cxsin(x)$ with real $Cnot=0.$ Indeed after multiplying both sides of the linear ODE by $Cxsin(x)$ we have
          $$Cxsin(x)left(frac {dy}{dx}+y(x)left(frac{1}{x}+cot(x)right)right)=Cxsin(x)$$
          that is
          $$Cfrac {d}{dx}left(xsin(x)y(x)right)=Cxsin(x)$$
          As you can see the value of the constant $Cnot =0$ is irrelevant.






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            $begingroup$

            No solution is lost here. It suffices to find ONE integrating factor. So $xsin(x)$ is enough. Nothing changes if we take $-xsin(x)$ or $Cxsin(x)$ with real $Cnot=0.$ Indeed after multiplying both sides of the linear ODE by $Cxsin(x)$ we have
            $$Cxsin(x)left(frac {dy}{dx}+y(x)left(frac{1}{x}+cot(x)right)right)=Cxsin(x)$$
            that is
            $$Cfrac {d}{dx}left(xsin(x)y(x)right)=Cxsin(x)$$
            As you can see the value of the constant $Cnot =0$ is irrelevant.






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              $begingroup$

              No solution is lost here. It suffices to find ONE integrating factor. So $xsin(x)$ is enough. Nothing changes if we take $-xsin(x)$ or $Cxsin(x)$ with real $Cnot=0.$ Indeed after multiplying both sides of the linear ODE by $Cxsin(x)$ we have
              $$Cxsin(x)left(frac {dy}{dx}+y(x)left(frac{1}{x}+cot(x)right)right)=Cxsin(x)$$
              that is
              $$Cfrac {d}{dx}left(xsin(x)y(x)right)=Cxsin(x)$$
              As you can see the value of the constant $Cnot =0$ is irrelevant.






              share|cite|improve this answer











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                $begingroup$

                No solution is lost here. It suffices to find ONE integrating factor. So $xsin(x)$ is enough. Nothing changes if we take $-xsin(x)$ or $Cxsin(x)$ with real $Cnot=0.$ Indeed after multiplying both sides of the linear ODE by $Cxsin(x)$ we have
                $$Cxsin(x)left(frac {dy}{dx}+y(x)left(frac{1}{x}+cot(x)right)right)=Cxsin(x)$$
                that is
                $$Cfrac {d}{dx}left(xsin(x)y(x)right)=Cxsin(x)$$
                As you can see the value of the constant $Cnot =0$ is irrelevant.






                share|cite|improve this answer











                $endgroup$



                No solution is lost here. It suffices to find ONE integrating factor. So $xsin(x)$ is enough. Nothing changes if we take $-xsin(x)$ or $Cxsin(x)$ with real $Cnot=0.$ Indeed after multiplying both sides of the linear ODE by $Cxsin(x)$ we have
                $$Cxsin(x)left(frac {dy}{dx}+y(x)left(frac{1}{x}+cot(x)right)right)=Cxsin(x)$$
                that is
                $$Cfrac {d}{dx}left(xsin(x)y(x)right)=Cxsin(x)$$
                As you can see the value of the constant $Cnot =0$ is irrelevant.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 18 at 17:37

























                answered Jan 18 at 17:23









                Robert ZRobert Z

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                102k1072145






























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