Differential equation with “backwards product rule”.












2














If we have the following differential equation, ($h,f$ known, $y$ unknown):



$$f'(x)y(x) + f(x)y'(x) = h(x)$$



it would be easy, since we could spot the derivative for a product:



$$(f(x)y(x))' = h(x)$$



and conclude



$$y(x) = frac{1}{f(x)}left(C + int h(x) dxright)$$



But what if we have it the "other way around", like this?



$$f(x)y(x) + f'(x)y'(x) = h(x)$$






real-analysis calculus differential-equations reference-request soft-question







share|cite|improve this question



























    2














    If we have the following differential equation, ($h,f$ known, $y$ unknown):



    $$f'(x)y(x) + f(x)y'(x) = h(x)$$



    it would be easy, since we could spot the derivative for a product:



    $$(f(x)y(x))' = h(x)$$



    and conclude



    $$y(x) = frac{1}{f(x)}left(C + int h(x) dxright)$$



    But what if we have it the "other way around", like this?



    $$f(x)y(x) + f'(x)y'(x) = h(x)$$






    real-analysis calculus differential-equations reference-request soft-question







    share|cite|improve this question

























      2












      2








      2


      4





      If we have the following differential equation, ($h,f$ known, $y$ unknown):



      $$f'(x)y(x) + f(x)y'(x) = h(x)$$



      it would be easy, since we could spot the derivative for a product:



      $$(f(x)y(x))' = h(x)$$



      and conclude



      $$y(x) = frac{1}{f(x)}left(C + int h(x) dxright)$$



      But what if we have it the "other way around", like this?



      $$f(x)y(x) + f'(x)y'(x) = h(x)$$






      real-analysis calculus differential-equations reference-request soft-question







      share|cite|improve this question













      If we have the following differential equation, ($h,f$ known, $y$ unknown):



      $$f'(x)y(x) + f(x)y'(x) = h(x)$$



      it would be easy, since we could spot the derivative for a product:



      $$(f(x)y(x))' = h(x)$$



      and conclude



      $$y(x) = frac{1}{f(x)}left(C + int h(x) dxright)$$



      But what if we have it the "other way around", like this?



      $$f(x)y(x) + f'(x)y'(x) = h(x)$$






      real-analysis calculus differential-equations reference-request soft-question




      real-analysis calculus differential-equations reference-request soft-question






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 at 7:30









      mathreadler

      14.7k72160




      14.7k72160






















          2 Answers
          2






          active

          oldest

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          1














          If the equation



          $f'(x)y(x) + f(x)y'(x) = h(x) tag 1$



          is written "the other way around",



          $f(x)y(x) + f'(x)y'(x) = h(x), tag 2$



          then there is no obvious way to apply the product rule; if we observe, however, that on an interval $J$ with



          $f'(x) ne 0, ; x in J, tag 3$



          then we may write (2) in the form



          $y'(x) + dfrac{f(x)}{f'(x)}y(x) = dfrac{h(x)}{f'(x)}, tag 4$



          which is a first-order system with varying coefficients, which has a well-known solution



          $y(x)$
          $= exp left ( -displaystyle int_{x_0}^x dfrac{f(s)}{f'(s)} ; ds right ) left (y(x_0) + exp left (displaystyle int_{x_0}^x dfrac{f(s)}{f'(s)} ; ds right) displaystyle int_{x_0}^x exp left ( -displaystyle int_{x_0}^s dfrac{f(u)}{f'(u)} ; du right ) dfrac{h(s)}{f'(s)} ; dsright ), tag 5$



          with $x_0 in J$.



          The formula (5) may in fact also be applied to (1) if we assume



          $f(x) ne 0, x in J, tag 6$



          and divide (1) by $f(x)$:



          $y'(x) + dfrac{f'(x)}{f(x)} y(x) = dfrac{h(x)}{f(x)}; tag 7$



          we obtain a formula which effectively interchanges $f(x)$ and $f'(x)$ in (5). It appears that the formula



          $ln f(x) - ln f(x_0) = displaystyle int_{x_0}^x dfrac{f'(s)}{f(s)} ; ds tag 8$



          may help further simplify (5) when applied to the case of (7).






          share|cite|improve this answer





























            2














            Let $h$ be an anti-derivative of $frac f {f'}$ Then $(e^{h}y)'=e^{h}(y'+frac f {f'} y)=e^{h}frac 1 {f'} (y'f'+fy)=e^{h}frac h {f'} $. Now integrate this.






            share|cite|improve this answer





















            • Ah so it is an integrating factor or (it was long time ago I did this)? Could you please expand a bit?
              – mathreadler
              Dec 10 at 7:50






            • 1




              The DE is of the form $y'+phi y=psi$ and it solved by multiplying the equation by the integrating factor $e^{int phi}$.
              – Kavi Rama Murthy
              Dec 10 at 7:52











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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            If the equation



            $f'(x)y(x) + f(x)y'(x) = h(x) tag 1$



            is written "the other way around",



            $f(x)y(x) + f'(x)y'(x) = h(x), tag 2$



            then there is no obvious way to apply the product rule; if we observe, however, that on an interval $J$ with



            $f'(x) ne 0, ; x in J, tag 3$



            then we may write (2) in the form



            $y'(x) + dfrac{f(x)}{f'(x)}y(x) = dfrac{h(x)}{f'(x)}, tag 4$



            which is a first-order system with varying coefficients, which has a well-known solution



            $y(x)$
            $= exp left ( -displaystyle int_{x_0}^x dfrac{f(s)}{f'(s)} ; ds right ) left (y(x_0) + exp left (displaystyle int_{x_0}^x dfrac{f(s)}{f'(s)} ; ds right) displaystyle int_{x_0}^x exp left ( -displaystyle int_{x_0}^s dfrac{f(u)}{f'(u)} ; du right ) dfrac{h(s)}{f'(s)} ; dsright ), tag 5$



            with $x_0 in J$.



            The formula (5) may in fact also be applied to (1) if we assume



            $f(x) ne 0, x in J, tag 6$



            and divide (1) by $f(x)$:



            $y'(x) + dfrac{f'(x)}{f(x)} y(x) = dfrac{h(x)}{f(x)}; tag 7$



            we obtain a formula which effectively interchanges $f(x)$ and $f'(x)$ in (5). It appears that the formula



            $ln f(x) - ln f(x_0) = displaystyle int_{x_0}^x dfrac{f'(s)}{f(s)} ; ds tag 8$



            may help further simplify (5) when applied to the case of (7).






            share|cite|improve this answer


























              1














              If the equation



              $f'(x)y(x) + f(x)y'(x) = h(x) tag 1$



              is written "the other way around",



              $f(x)y(x) + f'(x)y'(x) = h(x), tag 2$



              then there is no obvious way to apply the product rule; if we observe, however, that on an interval $J$ with



              $f'(x) ne 0, ; x in J, tag 3$



              then we may write (2) in the form



              $y'(x) + dfrac{f(x)}{f'(x)}y(x) = dfrac{h(x)}{f'(x)}, tag 4$



              which is a first-order system with varying coefficients, which has a well-known solution



              $y(x)$
              $= exp left ( -displaystyle int_{x_0}^x dfrac{f(s)}{f'(s)} ; ds right ) left (y(x_0) + exp left (displaystyle int_{x_0}^x dfrac{f(s)}{f'(s)} ; ds right) displaystyle int_{x_0}^x exp left ( -displaystyle int_{x_0}^s dfrac{f(u)}{f'(u)} ; du right ) dfrac{h(s)}{f'(s)} ; dsright ), tag 5$



              with $x_0 in J$.



              The formula (5) may in fact also be applied to (1) if we assume



              $f(x) ne 0, x in J, tag 6$



              and divide (1) by $f(x)$:



              $y'(x) + dfrac{f'(x)}{f(x)} y(x) = dfrac{h(x)}{f(x)}; tag 7$



              we obtain a formula which effectively interchanges $f(x)$ and $f'(x)$ in (5). It appears that the formula



              $ln f(x) - ln f(x_0) = displaystyle int_{x_0}^x dfrac{f'(s)}{f(s)} ; ds tag 8$



              may help further simplify (5) when applied to the case of (7).






              share|cite|improve this answer
























                1












                1








                1






                If the equation



                $f'(x)y(x) + f(x)y'(x) = h(x) tag 1$



                is written "the other way around",



                $f(x)y(x) + f'(x)y'(x) = h(x), tag 2$



                then there is no obvious way to apply the product rule; if we observe, however, that on an interval $J$ with



                $f'(x) ne 0, ; x in J, tag 3$



                then we may write (2) in the form



                $y'(x) + dfrac{f(x)}{f'(x)}y(x) = dfrac{h(x)}{f'(x)}, tag 4$



                which is a first-order system with varying coefficients, which has a well-known solution



                $y(x)$
                $= exp left ( -displaystyle int_{x_0}^x dfrac{f(s)}{f'(s)} ; ds right ) left (y(x_0) + exp left (displaystyle int_{x_0}^x dfrac{f(s)}{f'(s)} ; ds right) displaystyle int_{x_0}^x exp left ( -displaystyle int_{x_0}^s dfrac{f(u)}{f'(u)} ; du right ) dfrac{h(s)}{f'(s)} ; dsright ), tag 5$



                with $x_0 in J$.



                The formula (5) may in fact also be applied to (1) if we assume



                $f(x) ne 0, x in J, tag 6$



                and divide (1) by $f(x)$:



                $y'(x) + dfrac{f'(x)}{f(x)} y(x) = dfrac{h(x)}{f(x)}; tag 7$



                we obtain a formula which effectively interchanges $f(x)$ and $f'(x)$ in (5). It appears that the formula



                $ln f(x) - ln f(x_0) = displaystyle int_{x_0}^x dfrac{f'(s)}{f(s)} ; ds tag 8$



                may help further simplify (5) when applied to the case of (7).






                share|cite|improve this answer












                If the equation



                $f'(x)y(x) + f(x)y'(x) = h(x) tag 1$



                is written "the other way around",



                $f(x)y(x) + f'(x)y'(x) = h(x), tag 2$



                then there is no obvious way to apply the product rule; if we observe, however, that on an interval $J$ with



                $f'(x) ne 0, ; x in J, tag 3$



                then we may write (2) in the form



                $y'(x) + dfrac{f(x)}{f'(x)}y(x) = dfrac{h(x)}{f'(x)}, tag 4$



                which is a first-order system with varying coefficients, which has a well-known solution



                $y(x)$
                $= exp left ( -displaystyle int_{x_0}^x dfrac{f(s)}{f'(s)} ; ds right ) left (y(x_0) + exp left (displaystyle int_{x_0}^x dfrac{f(s)}{f'(s)} ; ds right) displaystyle int_{x_0}^x exp left ( -displaystyle int_{x_0}^s dfrac{f(u)}{f'(u)} ; du right ) dfrac{h(s)}{f'(s)} ; dsright ), tag 5$



                with $x_0 in J$.



                The formula (5) may in fact also be applied to (1) if we assume



                $f(x) ne 0, x in J, tag 6$



                and divide (1) by $f(x)$:



                $y'(x) + dfrac{f'(x)}{f(x)} y(x) = dfrac{h(x)}{f(x)}; tag 7$



                we obtain a formula which effectively interchanges $f(x)$ and $f'(x)$ in (5). It appears that the formula



                $ln f(x) - ln f(x_0) = displaystyle int_{x_0}^x dfrac{f'(s)}{f(s)} ; ds tag 8$



                may help further simplify (5) when applied to the case of (7).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 10 at 8:16









                Robert Lewis

                43.6k22863




                43.6k22863























                    2














                    Let $h$ be an anti-derivative of $frac f {f'}$ Then $(e^{h}y)'=e^{h}(y'+frac f {f'} y)=e^{h}frac 1 {f'} (y'f'+fy)=e^{h}frac h {f'} $. Now integrate this.






                    share|cite|improve this answer





















                    • Ah so it is an integrating factor or (it was long time ago I did this)? Could you please expand a bit?
                      – mathreadler
                      Dec 10 at 7:50






                    • 1




                      The DE is of the form $y'+phi y=psi$ and it solved by multiplying the equation by the integrating factor $e^{int phi}$.
                      – Kavi Rama Murthy
                      Dec 10 at 7:52
















                    2














                    Let $h$ be an anti-derivative of $frac f {f'}$ Then $(e^{h}y)'=e^{h}(y'+frac f {f'} y)=e^{h}frac 1 {f'} (y'f'+fy)=e^{h}frac h {f'} $. Now integrate this.






                    share|cite|improve this answer





















                    • Ah so it is an integrating factor or (it was long time ago I did this)? Could you please expand a bit?
                      – mathreadler
                      Dec 10 at 7:50






                    • 1




                      The DE is of the form $y'+phi y=psi$ and it solved by multiplying the equation by the integrating factor $e^{int phi}$.
                      – Kavi Rama Murthy
                      Dec 10 at 7:52














                    2












                    2








                    2






                    Let $h$ be an anti-derivative of $frac f {f'}$ Then $(e^{h}y)'=e^{h}(y'+frac f {f'} y)=e^{h}frac 1 {f'} (y'f'+fy)=e^{h}frac h {f'} $. Now integrate this.






                    share|cite|improve this answer












                    Let $h$ be an anti-derivative of $frac f {f'}$ Then $(e^{h}y)'=e^{h}(y'+frac f {f'} y)=e^{h}frac 1 {f'} (y'f'+fy)=e^{h}frac h {f'} $. Now integrate this.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 10 at 7:45









                    Kavi Rama Murthy

                    49.8k31854




                    49.8k31854












                    • Ah so it is an integrating factor or (it was long time ago I did this)? Could you please expand a bit?
                      – mathreadler
                      Dec 10 at 7:50






                    • 1




                      The DE is of the form $y'+phi y=psi$ and it solved by multiplying the equation by the integrating factor $e^{int phi}$.
                      – Kavi Rama Murthy
                      Dec 10 at 7:52


















                    • Ah so it is an integrating factor or (it was long time ago I did this)? Could you please expand a bit?
                      – mathreadler
                      Dec 10 at 7:50






                    • 1




                      The DE is of the form $y'+phi y=psi$ and it solved by multiplying the equation by the integrating factor $e^{int phi}$.
                      – Kavi Rama Murthy
                      Dec 10 at 7:52
















                    Ah so it is an integrating factor or (it was long time ago I did this)? Could you please expand a bit?
                    – mathreadler
                    Dec 10 at 7:50




                    Ah so it is an integrating factor or (it was long time ago I did this)? Could you please expand a bit?
                    – mathreadler
                    Dec 10 at 7:50




                    1




                    1




                    The DE is of the form $y'+phi y=psi$ and it solved by multiplying the equation by the integrating factor $e^{int phi}$.
                    – Kavi Rama Murthy
                    Dec 10 at 7:52




                    The DE is of the form $y'+phi y=psi$ and it solved by multiplying the equation by the integrating factor $e^{int phi}$.
                    – Kavi Rama Murthy
                    Dec 10 at 7:52


















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