Solving $frac{mathrm dy}{mathrm dx}+frac y{x^2}=frac1{x^2}$












1












$begingroup$


Solve the following equation:
$$
dfrac {dy}{dx} + dfrac {y}{x^2}=dfrac {1}{x^2}
$$



My Attempt



Given:
$$
dfrac {dy}{dx}+dfrac {y}{x^2}=dfrac {1}{x^2}
$$
Comparing above equation with the standard form of first order linear differential equation $dy/dx+P.y=Q$ where $P$ and $Q$ are the functions of $x$ or constants.



Now, using the integrating factor $e^{int P dx}$:
$$
e^{int P dx}=e^{int x^{-2} dx}=e^{-frac {1}{x}}
$$
Multiplying both sides of the given equation by integrating factor:
$$
e^{-frac {1}{x}}.dfrac {dy}{dx}+dfrac {y.e^{-frac {1}{x}}}{x^2}=dfrac {e^{-frac {1}{x}}}{x^2}
$$






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
    $endgroup$
    – John Hughes
    Dec 26 '18 at 13:18






  • 5




    $begingroup$
    @blue_eyed_: This is a Separable Equation.
    $endgroup$
    – Moo
    Dec 26 '18 at 13:20
















1












$begingroup$


Solve the following equation:
$$
dfrac {dy}{dx} + dfrac {y}{x^2}=dfrac {1}{x^2}
$$



My Attempt



Given:
$$
dfrac {dy}{dx}+dfrac {y}{x^2}=dfrac {1}{x^2}
$$
Comparing above equation with the standard form of first order linear differential equation $dy/dx+P.y=Q$ where $P$ and $Q$ are the functions of $x$ or constants.



Now, using the integrating factor $e^{int P dx}$:
$$
e^{int P dx}=e^{int x^{-2} dx}=e^{-frac {1}{x}}
$$
Multiplying both sides of the given equation by integrating factor:
$$
e^{-frac {1}{x}}.dfrac {dy}{dx}+dfrac {y.e^{-frac {1}{x}}}{x^2}=dfrac {e^{-frac {1}{x}}}{x^2}
$$






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
    $endgroup$
    – John Hughes
    Dec 26 '18 at 13:18






  • 5




    $begingroup$
    @blue_eyed_: This is a Separable Equation.
    $endgroup$
    – Moo
    Dec 26 '18 at 13:20














1












1








1





$begingroup$


Solve the following equation:
$$
dfrac {dy}{dx} + dfrac {y}{x^2}=dfrac {1}{x^2}
$$



My Attempt



Given:
$$
dfrac {dy}{dx}+dfrac {y}{x^2}=dfrac {1}{x^2}
$$
Comparing above equation with the standard form of first order linear differential equation $dy/dx+P.y=Q$ where $P$ and $Q$ are the functions of $x$ or constants.



Now, using the integrating factor $e^{int P dx}$:
$$
e^{int P dx}=e^{int x^{-2} dx}=e^{-frac {1}{x}}
$$
Multiplying both sides of the given equation by integrating factor:
$$
e^{-frac {1}{x}}.dfrac {dy}{dx}+dfrac {y.e^{-frac {1}{x}}}{x^2}=dfrac {e^{-frac {1}{x}}}{x^2}
$$






ordinary-differential-equations







share|cite|improve this question











$endgroup$




Solve the following equation:
$$
dfrac {dy}{dx} + dfrac {y}{x^2}=dfrac {1}{x^2}
$$



My Attempt



Given:
$$
dfrac {dy}{dx}+dfrac {y}{x^2}=dfrac {1}{x^2}
$$
Comparing above equation with the standard form of first order linear differential equation $dy/dx+P.y=Q$ where $P$ and $Q$ are the functions of $x$ or constants.



Now, using the integrating factor $e^{int P dx}$:
$$
e^{int P dx}=e^{int x^{-2} dx}=e^{-frac {1}{x}}
$$
Multiplying both sides of the given equation by integrating factor:
$$
e^{-frac {1}{x}}.dfrac {dy}{dx}+dfrac {y.e^{-frac {1}{x}}}{x^2}=dfrac {e^{-frac {1}{x}}}{x^2}
$$






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













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edited Dec 26 '18 at 15:16









Saad

19.7k92352




19.7k92352










asked Dec 26 '18 at 13:14









blue_eyed_...blue_eyed_...

3,26721750




3,26721750








  • 6




    $begingroup$
    OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
    $endgroup$
    – John Hughes
    Dec 26 '18 at 13:18






  • 5




    $begingroup$
    @blue_eyed_: This is a Separable Equation.
    $endgroup$
    – Moo
    Dec 26 '18 at 13:20














  • 6




    $begingroup$
    OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
    $endgroup$
    – John Hughes
    Dec 26 '18 at 13:18






  • 5




    $begingroup$
    @blue_eyed_: This is a Separable Equation.
    $endgroup$
    – Moo
    Dec 26 '18 at 13:20








6




6




$begingroup$
OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
$endgroup$
– John Hughes
Dec 26 '18 at 13:18




$begingroup$
OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
$endgroup$
– John Hughes
Dec 26 '18 at 13:18




5




5




$begingroup$
@blue_eyed_: This is a Separable Equation.
$endgroup$
– Moo
Dec 26 '18 at 13:20




$begingroup$
@blue_eyed_: This is a Separable Equation.
$endgroup$
– Moo
Dec 26 '18 at 13:20










4 Answers
4






active

oldest

votes


















4












$begingroup$

Hint: Write your equation in the form
$$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Great, now write it in the form
    $$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$



    We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.



    Hence $$frac{partial F}{partial y} = e^{-1/x}$$
    $$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$



    Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.



    Hence our equation is $$dF(x,y) = 0$$



    so $F(x,y) = C$ for some $C in mathbb{R}$. We get



    $$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        This is a linear DE so considering instead



        $$
        x^2y'+y=1
        $$



        $$
        y = y_h + y_p
        $$



        such that



        $$
        x^2y_h'+y_h = 0\
        x^2y_p' + y_p = 1
        $$



        for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives



        $$
        -C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
        $$



        and for the particular $y_p = 1$ then



        $$
        y = C_0 e^{frac 1x}+1
        $$






        share|cite|improve this answer











        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Hint: Write your equation in the form
          $$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$






          share|cite|improve this answer









          $endgroup$


















            4












            $begingroup$

            Hint: Write your equation in the form
            $$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$






            share|cite|improve this answer









            $endgroup$
















              4












              4








              4





              $begingroup$

              Hint: Write your equation in the form
              $$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$






              share|cite|improve this answer









              $endgroup$



              Hint: Write your equation in the form
              $$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 26 '18 at 13:29









              Dr. Sonnhard GraubnerDr. Sonnhard Graubner

              76.2k42866




              76.2k42866























                  2












                  $begingroup$

                  Great, now write it in the form
                  $$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$



                  We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.



                  Hence $$frac{partial F}{partial y} = e^{-1/x}$$
                  $$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$



                  Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.



                  Hence our equation is $$dF(x,y) = 0$$



                  so $F(x,y) = C$ for some $C in mathbb{R}$. We get



                  $$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    Great, now write it in the form
                    $$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$



                    We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.



                    Hence $$frac{partial F}{partial y} = e^{-1/x}$$
                    $$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$



                    Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.



                    Hence our equation is $$dF(x,y) = 0$$



                    so $F(x,y) = C$ for some $C in mathbb{R}$. We get



                    $$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      Great, now write it in the form
                      $$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$



                      We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.



                      Hence $$frac{partial F}{partial y} = e^{-1/x}$$
                      $$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$



                      Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.



                      Hence our equation is $$dF(x,y) = 0$$



                      so $F(x,y) = C$ for some $C in mathbb{R}$. We get



                      $$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$






                      share|cite|improve this answer









                      $endgroup$



                      Great, now write it in the form
                      $$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$



                      We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.



                      Hence $$frac{partial F}{partial y} = e^{-1/x}$$
                      $$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$



                      Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.



                      Hence our equation is $$dF(x,y) = 0$$



                      so $F(x,y) = C$ for some $C in mathbb{R}$. We get



                      $$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 26 '18 at 13:33









                      mechanodroidmechanodroid

                      27.8k62447




                      27.8k62447























                          1












                          $begingroup$

                          I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.






                              share|cite|improve this answer









                              $endgroup$



                              I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 26 '18 at 13:44









                              J.G.J.G.

                              27.8k22843




                              27.8k22843























                                  1












                                  $begingroup$

                                  This is a linear DE so considering instead



                                  $$
                                  x^2y'+y=1
                                  $$



                                  $$
                                  y = y_h + y_p
                                  $$



                                  such that



                                  $$
                                  x^2y_h'+y_h = 0\
                                  x^2y_p' + y_p = 1
                                  $$



                                  for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives



                                  $$
                                  -C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
                                  $$



                                  and for the particular $y_p = 1$ then



                                  $$
                                  y = C_0 e^{frac 1x}+1
                                  $$






                                  share|cite|improve this answer











                                  $endgroup$


















                                    1












                                    $begingroup$

                                    This is a linear DE so considering instead



                                    $$
                                    x^2y'+y=1
                                    $$



                                    $$
                                    y = y_h + y_p
                                    $$



                                    such that



                                    $$
                                    x^2y_h'+y_h = 0\
                                    x^2y_p' + y_p = 1
                                    $$



                                    for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives



                                    $$
                                    -C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
                                    $$



                                    and for the particular $y_p = 1$ then



                                    $$
                                    y = C_0 e^{frac 1x}+1
                                    $$






                                    share|cite|improve this answer











                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      This is a linear DE so considering instead



                                      $$
                                      x^2y'+y=1
                                      $$



                                      $$
                                      y = y_h + y_p
                                      $$



                                      such that



                                      $$
                                      x^2y_h'+y_h = 0\
                                      x^2y_p' + y_p = 1
                                      $$



                                      for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives



                                      $$
                                      -C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
                                      $$



                                      and for the particular $y_p = 1$ then



                                      $$
                                      y = C_0 e^{frac 1x}+1
                                      $$






                                      share|cite|improve this answer











                                      $endgroup$



                                      This is a linear DE so considering instead



                                      $$
                                      x^2y'+y=1
                                      $$



                                      $$
                                      y = y_h + y_p
                                      $$



                                      such that



                                      $$
                                      x^2y_h'+y_h = 0\
                                      x^2y_p' + y_p = 1
                                      $$



                                      for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives



                                      $$
                                      -C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
                                      $$



                                      and for the particular $y_p = 1$ then



                                      $$
                                      y = C_0 e^{frac 1x}+1
                                      $$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Dec 26 '18 at 14:53

























                                      answered Dec 26 '18 at 13:50









                                      CesareoCesareo

                                      8,9913516




                                      8,9913516






























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                                          Mont Emei

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                                          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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