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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share|cite|improve this question








edited Dec 26 '18 at 15:16









Saad

19.7k92352




19.7k92352










asked Dec 26 '18 at 13:14









blue_eyed_...blue_eyed_...

3,25921649




3,25921649








  • 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

              75.3k42865




              75.3k42865























                  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

                      27.6k62447




                      27.6k62447























                          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.

                              26.3k22541




                              26.3k22541























                                  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

                                      8,8293516




                                      8,8293516






























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                                          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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