Solving Integral Equation by Converting to Differential Equations The Next CEO of Stack OverflowAre there methods to solve coupled integral and integro-differential equations?Voltera equationSolve integral equation by converting to differential equationHow can I solve this integral equation by converting it to a differential equationConverting a integral equation to differential equationSolving integro-differential equation - numericallySolution of Differential equation as an integral equationConverting Differential Operator to Integral Equationreference for converting an integro-differential equation to a differential algebraic equationSolving second order ordinary differential equation with variable constants

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Solving Integral Equation by Converting to Differential Equations



The Next CEO of Stack OverflowAre there methods to solve coupled integral and integro-differential equations?Voltera equationSolve integral equation by converting to differential equationHow can I solve this integral equation by converting it to a differential equationConverting a integral equation to differential equationSolving integro-differential equation - numericallySolution of Differential equation as an integral equationConverting Differential Operator to Integral Equationreference for converting an integro-differential equation to a differential algebraic equationSolving second order ordinary differential equation with variable constants










2












$begingroup$


Consider the problem



$$phi(x) = x - int_0^x(x-s)phi(s),ds$$



How can we solve this by converting to a differential equation?










share|cite|improve this question









$endgroup$
















    2












    $begingroup$


    Consider the problem



    $$phi(x) = x - int_0^x(x-s)phi(s),ds$$



    How can we solve this by converting to a differential equation?










    share|cite|improve this question









    $endgroup$














      2












      2








      2





      $begingroup$


      Consider the problem



      $$phi(x) = x - int_0^x(x-s)phi(s),ds$$



      How can we solve this by converting to a differential equation?










      share|cite|improve this question









      $endgroup$




      Consider the problem



      $$phi(x) = x - int_0^x(x-s)phi(s),ds$$



      How can we solve this by converting to a differential equation?







      ordinary-differential-equations integral-equations integro-differential-equations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 5 hours ago









      LightningStrikeLightningStrike

      555




      555




















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          We have that
          $$phi(x)=x-xint_0^x phi(s) mathrmd s + int_0^x s phi(s)mathrmds$$
          From this, we can see that $phi(0)=0$.
          We can differentiate both sides and use the product rule and the FTC1 to get:
          $$phi'(x)=1-int_0^x phi(s) mathrmds -x phi(x)+xphi(x)$$
          $$phi'(x)=1-int_0^x phi(s) mathrmd s$$
          From this, we can see that $phi'(0)=1$. We can differentiate it again:
          $$phi''(x)=-phi(x)$$
          Which is an alternative definition of the $sin$ function.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            In fact, the only valid solution for $phi(x)$ is $sin(x)$ because of the original equation.
            $endgroup$
            – Peter Foreman
            4 hours ago










          • $begingroup$
            @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
            $endgroup$
            – Botond
            4 hours ago











          • $begingroup$
            Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
            $endgroup$
            – LightningStrike
            4 hours ago










          • $begingroup$
            @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrmds$?
            $endgroup$
            – Botond
            4 hours ago


















          1












          $begingroup$

          Differentiating both sides using Leibniz rule :



          $$phi '(x)=1-int_0^xphi (s)ds$$



          Differentiate again:



          $$phi ''(x)=-phi (x)$$






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
            $endgroup$
            – Botond
            4 hours ago










          • $begingroup$
            may be you are right...but this is a common technique in an introductory course of integral equations.
            $endgroup$
            – logo
            4 hours ago











          • $begingroup$
            I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
            $endgroup$
            – Botond
            3 hours ago












          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          We have that
          $$phi(x)=x-xint_0^x phi(s) mathrmd s + int_0^x s phi(s)mathrmds$$
          From this, we can see that $phi(0)=0$.
          We can differentiate both sides and use the product rule and the FTC1 to get:
          $$phi'(x)=1-int_0^x phi(s) mathrmds -x phi(x)+xphi(x)$$
          $$phi'(x)=1-int_0^x phi(s) mathrmd s$$
          From this, we can see that $phi'(0)=1$. We can differentiate it again:
          $$phi''(x)=-phi(x)$$
          Which is an alternative definition of the $sin$ function.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            In fact, the only valid solution for $phi(x)$ is $sin(x)$ because of the original equation.
            $endgroup$
            – Peter Foreman
            4 hours ago










          • $begingroup$
            @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
            $endgroup$
            – Botond
            4 hours ago











          • $begingroup$
            Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
            $endgroup$
            – LightningStrike
            4 hours ago










          • $begingroup$
            @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrmds$?
            $endgroup$
            – Botond
            4 hours ago















          4












          $begingroup$

          We have that
          $$phi(x)=x-xint_0^x phi(s) mathrmd s + int_0^x s phi(s)mathrmds$$
          From this, we can see that $phi(0)=0$.
          We can differentiate both sides and use the product rule and the FTC1 to get:
          $$phi'(x)=1-int_0^x phi(s) mathrmds -x phi(x)+xphi(x)$$
          $$phi'(x)=1-int_0^x phi(s) mathrmd s$$
          From this, we can see that $phi'(0)=1$. We can differentiate it again:
          $$phi''(x)=-phi(x)$$
          Which is an alternative definition of the $sin$ function.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            In fact, the only valid solution for $phi(x)$ is $sin(x)$ because of the original equation.
            $endgroup$
            – Peter Foreman
            4 hours ago










          • $begingroup$
            @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
            $endgroup$
            – Botond
            4 hours ago











          • $begingroup$
            Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
            $endgroup$
            – LightningStrike
            4 hours ago










          • $begingroup$
            @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrmds$?
            $endgroup$
            – Botond
            4 hours ago













          4












          4








          4





          $begingroup$

          We have that
          $$phi(x)=x-xint_0^x phi(s) mathrmd s + int_0^x s phi(s)mathrmds$$
          From this, we can see that $phi(0)=0$.
          We can differentiate both sides and use the product rule and the FTC1 to get:
          $$phi'(x)=1-int_0^x phi(s) mathrmds -x phi(x)+xphi(x)$$
          $$phi'(x)=1-int_0^x phi(s) mathrmd s$$
          From this, we can see that $phi'(0)=1$. We can differentiate it again:
          $$phi''(x)=-phi(x)$$
          Which is an alternative definition of the $sin$ function.






          share|cite|improve this answer











          $endgroup$



          We have that
          $$phi(x)=x-xint_0^x phi(s) mathrmd s + int_0^x s phi(s)mathrmds$$
          From this, we can see that $phi(0)=0$.
          We can differentiate both sides and use the product rule and the FTC1 to get:
          $$phi'(x)=1-int_0^x phi(s) mathrmds -x phi(x)+xphi(x)$$
          $$phi'(x)=1-int_0^x phi(s) mathrmd s$$
          From this, we can see that $phi'(0)=1$. We can differentiate it again:
          $$phi''(x)=-phi(x)$$
          Which is an alternative definition of the $sin$ function.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 5 hours ago

























          answered 5 hours ago









          BotondBotond

          6,49331034




          6,49331034











          • $begingroup$
            In fact, the only valid solution for $phi(x)$ is $sin(x)$ because of the original equation.
            $endgroup$
            – Peter Foreman
            4 hours ago










          • $begingroup$
            @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
            $endgroup$
            – Botond
            4 hours ago











          • $begingroup$
            Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
            $endgroup$
            – LightningStrike
            4 hours ago










          • $begingroup$
            @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrmds$?
            $endgroup$
            – Botond
            4 hours ago
















          • $begingroup$
            In fact, the only valid solution for $phi(x)$ is $sin(x)$ because of the original equation.
            $endgroup$
            – Peter Foreman
            4 hours ago










          • $begingroup$
            @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
            $endgroup$
            – Botond
            4 hours ago











          • $begingroup$
            Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
            $endgroup$
            – LightningStrike
            4 hours ago










          • $begingroup$
            @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrmds$?
            $endgroup$
            – Botond
            4 hours ago















          $begingroup$
          In fact, the only valid solution for $phi(x)$ is $sin(x)$ because of the original equation.
          $endgroup$
          – Peter Foreman
          4 hours ago




          $begingroup$
          In fact, the only valid solution for $phi(x)$ is $sin(x)$ because of the original equation.
          $endgroup$
          – Peter Foreman
          4 hours ago












          $begingroup$
          @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
          $endgroup$
          – Botond
          4 hours ago





          $begingroup$
          @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
          $endgroup$
          – Botond
          4 hours ago













          $begingroup$
          Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
          $endgroup$
          – LightningStrike
          4 hours ago




          $begingroup$
          Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
          $endgroup$
          – LightningStrike
          4 hours ago












          $begingroup$
          @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrmds$?
          $endgroup$
          – Botond
          4 hours ago




          $begingroup$
          @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrmds$?
          $endgroup$
          – Botond
          4 hours ago











          1












          $begingroup$

          Differentiating both sides using Leibniz rule :



          $$phi '(x)=1-int_0^xphi (s)ds$$



          Differentiate again:



          $$phi ''(x)=-phi (x)$$






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
            $endgroup$
            – Botond
            4 hours ago










          • $begingroup$
            may be you are right...but this is a common technique in an introductory course of integral equations.
            $endgroup$
            – logo
            4 hours ago











          • $begingroup$
            I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
            $endgroup$
            – Botond
            3 hours ago
















          1












          $begingroup$

          Differentiating both sides using Leibniz rule :



          $$phi '(x)=1-int_0^xphi (s)ds$$



          Differentiate again:



          $$phi ''(x)=-phi (x)$$






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
            $endgroup$
            – Botond
            4 hours ago










          • $begingroup$
            may be you are right...but this is a common technique in an introductory course of integral equations.
            $endgroup$
            – logo
            4 hours ago











          • $begingroup$
            I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
            $endgroup$
            – Botond
            3 hours ago














          1












          1








          1





          $begingroup$

          Differentiating both sides using Leibniz rule :



          $$phi '(x)=1-int_0^xphi (s)ds$$



          Differentiate again:



          $$phi ''(x)=-phi (x)$$






          share|cite|improve this answer











          $endgroup$



          Differentiating both sides using Leibniz rule :



          $$phi '(x)=1-int_0^xphi (s)ds$$



          Differentiate again:



          $$phi ''(x)=-phi (x)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 5 hours ago

























          answered 5 hours ago









          logologo

          1048




          1048







          • 1




            $begingroup$
            Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
            $endgroup$
            – Botond
            4 hours ago










          • $begingroup$
            may be you are right...but this is a common technique in an introductory course of integral equations.
            $endgroup$
            – logo
            4 hours ago











          • $begingroup$
            I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
            $endgroup$
            – Botond
            3 hours ago













          • 1




            $begingroup$
            Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
            $endgroup$
            – Botond
            4 hours ago










          • $begingroup$
            may be you are right...but this is a common technique in an introductory course of integral equations.
            $endgroup$
            – logo
            4 hours ago











          • $begingroup$
            I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
            $endgroup$
            – Botond
            3 hours ago








          1




          1




          $begingroup$
          Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
          $endgroup$
          – Botond
          4 hours ago




          $begingroup$
          Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
          $endgroup$
          – Botond
          4 hours ago












          $begingroup$
          may be you are right...but this is a common technique in an introductory course of integral equations.
          $endgroup$
          – logo
          4 hours ago





          $begingroup$
          may be you are right...but this is a common technique in an introductory course of integral equations.
          $endgroup$
          – logo
          4 hours ago













          $begingroup$
          I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
          $endgroup$
          – Botond
          3 hours ago





          $begingroup$
          I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
          $endgroup$
          – Botond
          3 hours ago


















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