How to solve this ode $xy''-(1+x)y'+y=x^2$?












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Find the general solution of $xy''-(1+x)y'+y=x^2$ knowing that the homogeneous equation has the following solution: $e^{ax}$, where $a$ is a parameter you have to find.



I have found that $a=1$ or $a=1/x$.






ordinary-differential-equations







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  • 1




    $begingroup$
    The parameter cannot depend on $x$.
    $endgroup$
    – dmtri
    Jan 3 at 17:04
















4












$begingroup$


Find the general solution of $xy''-(1+x)y'+y=x^2$ knowing that the homogeneous equation has the following solution: $e^{ax}$, where $a$ is a parameter you have to find.



I have found that $a=1$ or $a=1/x$.






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The parameter cannot depend on $x$.
    $endgroup$
    – dmtri
    Jan 3 at 17:04














4












4








4


2



$begingroup$


Find the general solution of $xy''-(1+x)y'+y=x^2$ knowing that the homogeneous equation has the following solution: $e^{ax}$, where $a$ is a parameter you have to find.



I have found that $a=1$ or $a=1/x$.






ordinary-differential-equations







share|cite|improve this question











$endgroup$




Find the general solution of $xy''-(1+x)y'+y=x^2$ knowing that the homogeneous equation has the following solution: $e^{ax}$, where $a$ is a parameter you have to find.



I have found that $a=1$ or $a=1/x$.






ordinary-differential-equations




ordinary-differential-equations






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share|cite|improve this question













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edited Jan 3 at 17:12









dmtri

1,5262521




1,5262521










asked Jan 3 at 16:57









Gloria GuerreroGloria Guerrero

211




211








  • 1




    $begingroup$
    The parameter cannot depend on $x$.
    $endgroup$
    – dmtri
    Jan 3 at 17:04














  • 1




    $begingroup$
    The parameter cannot depend on $x$.
    $endgroup$
    – dmtri
    Jan 3 at 17:04








1




1




$begingroup$
The parameter cannot depend on $x$.
$endgroup$
– dmtri
Jan 3 at 17:04




$begingroup$
The parameter cannot depend on $x$.
$endgroup$
– dmtri
Jan 3 at 17:04










3 Answers
3






active

oldest

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1












$begingroup$

Hint. You are correct, the homogeneous case of the given linear ODE has $e^x$ as a solution (the parameter $a$ is supposed to be a constant real number). Note that $y=-x^2$ is a particular solution. Is there any other polynomial solution? Substitute $y(x)=Ax^2+Bx+C$ into the ODE, then
$$x(2A)-(1+x)(2Ax+B)+(Ax^2+Bx+C)=x^2$$
What may we conclude?






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    The equation $xy''-(1+x)y'+y=x^2$ can be rewritten as $(xy''-y')+(y-xy')=x^2.$ After division by $x^2$ we have
    $$
    frac{xy''-y'}{x^2} + frac{y-xy'}{x^2} = 1,
    $$
    which can be rewritten as
    $$
    Big( frac{y'}{x} Big)' - Big( frac{y}{x} Big)' = 1.
    $$



    Taking the antiderivative gives
    $$
    frac{y'}{x} - frac{y}{x} = x + C,
    $$
    which after multiplication with $x$ gives
    $$y' - y = x^2+ Cx.$$



    Now we multiply with the integrating factor $e^{-x}$:
    $$e^{-x} y' - e^{-x} y = x^2 e^{-x} + Cx e^{-x}.$$



    The left hand side can be rewritten as a derivative:
    $$( e^{-x} y )' = x^2 e^{-x} + Cx e^{-x}.$$



    Taking the antiderivative gives
    $$e^{-x} y = -(x^2+2x+2) e^{-x} - C(x+1) e^{-x} + D.$$



    Thus, all solutions to the original differential equation are given by
    $$y = -(x^2+2x+2) - C(x+1) + D e^x,$$
    where $C$ and $D$ are two constants.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Substitute $$y(x)=(-x-1)v(x)$$ then we get
      $$-xfrac{d^2v(x)}{dx^2(x+1)}+frac{dv(x)}{dx}(x^2+1)=x^2$$
      let $$frac{dv(x)}{dx}=u(x)$$ then we get
      $$frac{du(x)}{dx}+frac{-x^2-1)u(x)}{x(x+1)}=-frac{x}{x+1}$$
      with $$mu(x)=e^{intfrac{-x^2-1}{x(x+1)}dx}=frac{e^{-x}(x+1)^2}{x}$$ we get
      $$frac{e^{-x}(x+1)^2}{x}frac{du(x)}{dx}+frac{d}{dx}left(frac{e^{-x}(x+1)^2}{x}right)u(x)=-e^{-x}(x+1)$$ and this is
      $$intfrac{d}{dx}left(frac{e^{-x}(x+1)^2u(x)}{x}right)=int-e^{-x}(x+1)dx$$
      Can you finish?






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

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        active

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        active

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        1












        $begingroup$

        Hint. You are correct, the homogeneous case of the given linear ODE has $e^x$ as a solution (the parameter $a$ is supposed to be a constant real number). Note that $y=-x^2$ is a particular solution. Is there any other polynomial solution? Substitute $y(x)=Ax^2+Bx+C$ into the ODE, then
        $$x(2A)-(1+x)(2Ax+B)+(Ax^2+Bx+C)=x^2$$
        What may we conclude?






        share|cite|improve this answer











        $endgroup$


















          1












          $begingroup$

          Hint. You are correct, the homogeneous case of the given linear ODE has $e^x$ as a solution (the parameter $a$ is supposed to be a constant real number). Note that $y=-x^2$ is a particular solution. Is there any other polynomial solution? Substitute $y(x)=Ax^2+Bx+C$ into the ODE, then
          $$x(2A)-(1+x)(2Ax+B)+(Ax^2+Bx+C)=x^2$$
          What may we conclude?






          share|cite|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

            Hint. You are correct, the homogeneous case of the given linear ODE has $e^x$ as a solution (the parameter $a$ is supposed to be a constant real number). Note that $y=-x^2$ is a particular solution. Is there any other polynomial solution? Substitute $y(x)=Ax^2+Bx+C$ into the ODE, then
            $$x(2A)-(1+x)(2Ax+B)+(Ax^2+Bx+C)=x^2$$
            What may we conclude?






            share|cite|improve this answer











            $endgroup$



            Hint. You are correct, the homogeneous case of the given linear ODE has $e^x$ as a solution (the parameter $a$ is supposed to be a constant real number). Note that $y=-x^2$ is a particular solution. Is there any other polynomial solution? Substitute $y(x)=Ax^2+Bx+C$ into the ODE, then
            $$x(2A)-(1+x)(2Ax+B)+(Ax^2+Bx+C)=x^2$$
            What may we conclude?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 3 at 17:28

























            answered Jan 3 at 17:05









            Robert ZRobert Z

            99.1k1068139




            99.1k1068139























                1












                $begingroup$

                The equation $xy''-(1+x)y'+y=x^2$ can be rewritten as $(xy''-y')+(y-xy')=x^2.$ After division by $x^2$ we have
                $$
                frac{xy''-y'}{x^2} + frac{y-xy'}{x^2} = 1,
                $$
                which can be rewritten as
                $$
                Big( frac{y'}{x} Big)' - Big( frac{y}{x} Big)' = 1.
                $$



                Taking the antiderivative gives
                $$
                frac{y'}{x} - frac{y}{x} = x + C,
                $$
                which after multiplication with $x$ gives
                $$y' - y = x^2+ Cx.$$



                Now we multiply with the integrating factor $e^{-x}$:
                $$e^{-x} y' - e^{-x} y = x^2 e^{-x} + Cx e^{-x}.$$



                The left hand side can be rewritten as a derivative:
                $$( e^{-x} y )' = x^2 e^{-x} + Cx e^{-x}.$$



                Taking the antiderivative gives
                $$e^{-x} y = -(x^2+2x+2) e^{-x} - C(x+1) e^{-x} + D.$$



                Thus, all solutions to the original differential equation are given by
                $$y = -(x^2+2x+2) - C(x+1) + D e^x,$$
                where $C$ and $D$ are two constants.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  The equation $xy''-(1+x)y'+y=x^2$ can be rewritten as $(xy''-y')+(y-xy')=x^2.$ After division by $x^2$ we have
                  $$
                  frac{xy''-y'}{x^2} + frac{y-xy'}{x^2} = 1,
                  $$
                  which can be rewritten as
                  $$
                  Big( frac{y'}{x} Big)' - Big( frac{y}{x} Big)' = 1.
                  $$



                  Taking the antiderivative gives
                  $$
                  frac{y'}{x} - frac{y}{x} = x + C,
                  $$
                  which after multiplication with $x$ gives
                  $$y' - y = x^2+ Cx.$$



                  Now we multiply with the integrating factor $e^{-x}$:
                  $$e^{-x} y' - e^{-x} y = x^2 e^{-x} + Cx e^{-x}.$$



                  The left hand side can be rewritten as a derivative:
                  $$( e^{-x} y )' = x^2 e^{-x} + Cx e^{-x}.$$



                  Taking the antiderivative gives
                  $$e^{-x} y = -(x^2+2x+2) e^{-x} - C(x+1) e^{-x} + D.$$



                  Thus, all solutions to the original differential equation are given by
                  $$y = -(x^2+2x+2) - C(x+1) + D e^x,$$
                  where $C$ and $D$ are two constants.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    The equation $xy''-(1+x)y'+y=x^2$ can be rewritten as $(xy''-y')+(y-xy')=x^2.$ After division by $x^2$ we have
                    $$
                    frac{xy''-y'}{x^2} + frac{y-xy'}{x^2} = 1,
                    $$
                    which can be rewritten as
                    $$
                    Big( frac{y'}{x} Big)' - Big( frac{y}{x} Big)' = 1.
                    $$



                    Taking the antiderivative gives
                    $$
                    frac{y'}{x} - frac{y}{x} = x + C,
                    $$
                    which after multiplication with $x$ gives
                    $$y' - y = x^2+ Cx.$$



                    Now we multiply with the integrating factor $e^{-x}$:
                    $$e^{-x} y' - e^{-x} y = x^2 e^{-x} + Cx e^{-x}.$$



                    The left hand side can be rewritten as a derivative:
                    $$( e^{-x} y )' = x^2 e^{-x} + Cx e^{-x}.$$



                    Taking the antiderivative gives
                    $$e^{-x} y = -(x^2+2x+2) e^{-x} - C(x+1) e^{-x} + D.$$



                    Thus, all solutions to the original differential equation are given by
                    $$y = -(x^2+2x+2) - C(x+1) + D e^x,$$
                    where $C$ and $D$ are two constants.






                    share|cite|improve this answer









                    $endgroup$



                    The equation $xy''-(1+x)y'+y=x^2$ can be rewritten as $(xy''-y')+(y-xy')=x^2.$ After division by $x^2$ we have
                    $$
                    frac{xy''-y'}{x^2} + frac{y-xy'}{x^2} = 1,
                    $$
                    which can be rewritten as
                    $$
                    Big( frac{y'}{x} Big)' - Big( frac{y}{x} Big)' = 1.
                    $$



                    Taking the antiderivative gives
                    $$
                    frac{y'}{x} - frac{y}{x} = x + C,
                    $$
                    which after multiplication with $x$ gives
                    $$y' - y = x^2+ Cx.$$



                    Now we multiply with the integrating factor $e^{-x}$:
                    $$e^{-x} y' - e^{-x} y = x^2 e^{-x} + Cx e^{-x}.$$



                    The left hand side can be rewritten as a derivative:
                    $$( e^{-x} y )' = x^2 e^{-x} + Cx e^{-x}.$$



                    Taking the antiderivative gives
                    $$e^{-x} y = -(x^2+2x+2) e^{-x} - C(x+1) e^{-x} + D.$$



                    Thus, all solutions to the original differential equation are given by
                    $$y = -(x^2+2x+2) - C(x+1) + D e^x,$$
                    where $C$ and $D$ are two constants.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 3 at 17:48









                    md2perpemd2perpe

                    8,11611028




                    8,11611028























                        0












                        $begingroup$

                        Substitute $$y(x)=(-x-1)v(x)$$ then we get
                        $$-xfrac{d^2v(x)}{dx^2(x+1)}+frac{dv(x)}{dx}(x^2+1)=x^2$$
                        let $$frac{dv(x)}{dx}=u(x)$$ then we get
                        $$frac{du(x)}{dx}+frac{-x^2-1)u(x)}{x(x+1)}=-frac{x}{x+1}$$
                        with $$mu(x)=e^{intfrac{-x^2-1}{x(x+1)}dx}=frac{e^{-x}(x+1)^2}{x}$$ we get
                        $$frac{e^{-x}(x+1)^2}{x}frac{du(x)}{dx}+frac{d}{dx}left(frac{e^{-x}(x+1)^2}{x}right)u(x)=-e^{-x}(x+1)$$ and this is
                        $$intfrac{d}{dx}left(frac{e^{-x}(x+1)^2u(x)}{x}right)=int-e^{-x}(x+1)dx$$
                        Can you finish?






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Substitute $$y(x)=(-x-1)v(x)$$ then we get
                          $$-xfrac{d^2v(x)}{dx^2(x+1)}+frac{dv(x)}{dx}(x^2+1)=x^2$$
                          let $$frac{dv(x)}{dx}=u(x)$$ then we get
                          $$frac{du(x)}{dx}+frac{-x^2-1)u(x)}{x(x+1)}=-frac{x}{x+1}$$
                          with $$mu(x)=e^{intfrac{-x^2-1}{x(x+1)}dx}=frac{e^{-x}(x+1)^2}{x}$$ we get
                          $$frac{e^{-x}(x+1)^2}{x}frac{du(x)}{dx}+frac{d}{dx}left(frac{e^{-x}(x+1)^2}{x}right)u(x)=-e^{-x}(x+1)$$ and this is
                          $$intfrac{d}{dx}left(frac{e^{-x}(x+1)^2u(x)}{x}right)=int-e^{-x}(x+1)dx$$
                          Can you finish?






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Substitute $$y(x)=(-x-1)v(x)$$ then we get
                            $$-xfrac{d^2v(x)}{dx^2(x+1)}+frac{dv(x)}{dx}(x^2+1)=x^2$$
                            let $$frac{dv(x)}{dx}=u(x)$$ then we get
                            $$frac{du(x)}{dx}+frac{-x^2-1)u(x)}{x(x+1)}=-frac{x}{x+1}$$
                            with $$mu(x)=e^{intfrac{-x^2-1}{x(x+1)}dx}=frac{e^{-x}(x+1)^2}{x}$$ we get
                            $$frac{e^{-x}(x+1)^2}{x}frac{du(x)}{dx}+frac{d}{dx}left(frac{e^{-x}(x+1)^2}{x}right)u(x)=-e^{-x}(x+1)$$ and this is
                            $$intfrac{d}{dx}left(frac{e^{-x}(x+1)^2u(x)}{x}right)=int-e^{-x}(x+1)dx$$
                            Can you finish?






                            share|cite|improve this answer









                            $endgroup$



                            Substitute $$y(x)=(-x-1)v(x)$$ then we get
                            $$-xfrac{d^2v(x)}{dx^2(x+1)}+frac{dv(x)}{dx}(x^2+1)=x^2$$
                            let $$frac{dv(x)}{dx}=u(x)$$ then we get
                            $$frac{du(x)}{dx}+frac{-x^2-1)u(x)}{x(x+1)}=-frac{x}{x+1}$$
                            with $$mu(x)=e^{intfrac{-x^2-1}{x(x+1)}dx}=frac{e^{-x}(x+1)^2}{x}$$ we get
                            $$frac{e^{-x}(x+1)^2}{x}frac{du(x)}{dx}+frac{d}{dx}left(frac{e^{-x}(x+1)^2}{x}right)u(x)=-e^{-x}(x+1)$$ and this is
                            $$intfrac{d}{dx}left(frac{e^{-x}(x+1)^2u(x)}{x}right)=int-e^{-x}(x+1)dx$$
                            Can you finish?







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 3 at 17:18









                            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                            76.8k42866




                            76.8k42866






























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