Verify the homogeneous differential equation.












0














$xy''-(1+x)y'+y=x^2e^{2x}$



How to verify that $y_1=e^x$ and $y_2=1+x$ are solution for homogeneous differential equation.



I already plugging $y_1=e^x,y_1'=e^x,y_1''=e^x$ and $y_2=1+x,y_2'=1,y_2''=0$ into equation but i get $x^2e^{2x}$ still didn't get $0$.






differential-equations







share|cite|improve this question
























  • is it $e^{2x}$ or $e^2cdot x$?
    – Dylan
    Nov 26 '18 at 14:07










  • opps, sorry i did some mistake is $e^{2x}$, i already edit
    – LTY
    Nov 26 '18 at 14:08


















0














$xy''-(1+x)y'+y=x^2e^{2x}$



How to verify that $y_1=e^x$ and $y_2=1+x$ are solution for homogeneous differential equation.



I already plugging $y_1=e^x,y_1'=e^x,y_1''=e^x$ and $y_2=1+x,y_2'=1,y_2''=0$ into equation but i get $x^2e^{2x}$ still didn't get $0$.






differential-equations







share|cite|improve this question
























  • is it $e^{2x}$ or $e^2cdot x$?
    – Dylan
    Nov 26 '18 at 14:07










  • opps, sorry i did some mistake is $e^{2x}$, i already edit
    – LTY
    Nov 26 '18 at 14:08
















0












0








0







$xy''-(1+x)y'+y=x^2e^{2x}$



How to verify that $y_1=e^x$ and $y_2=1+x$ are solution for homogeneous differential equation.



I already plugging $y_1=e^x,y_1'=e^x,y_1''=e^x$ and $y_2=1+x,y_2'=1,y_2''=0$ into equation but i get $x^2e^{2x}$ still didn't get $0$.






differential-equations







share|cite|improve this question















$xy''-(1+x)y'+y=x^2e^{2x}$



How to verify that $y_1=e^x$ and $y_2=1+x$ are solution for homogeneous differential equation.



I already plugging $y_1=e^x,y_1'=e^x,y_1''=e^x$ and $y_2=1+x,y_2'=1,y_2''=0$ into equation but i get $x^2e^{2x}$ still didn't get $0$.






differential-equations




differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 14:10

























asked Nov 26 '18 at 13:20









LTY

285




285












  • is it $e^{2x}$ or $e^2cdot x$?
    – Dylan
    Nov 26 '18 at 14:07










  • opps, sorry i did some mistake is $e^{2x}$, i already edit
    – LTY
    Nov 26 '18 at 14:08




















  • is it $e^{2x}$ or $e^2cdot x$?
    – Dylan
    Nov 26 '18 at 14:07










  • opps, sorry i did some mistake is $e^{2x}$, i already edit
    – LTY
    Nov 26 '18 at 14:08


















is it $e^{2x}$ or $e^2cdot x$?
– Dylan
Nov 26 '18 at 14:07




is it $e^{2x}$ or $e^2cdot x$?
– Dylan
Nov 26 '18 at 14:07












opps, sorry i did some mistake is $e^{2x}$, i already edit
– LTY
Nov 26 '18 at 14:08






opps, sorry i did some mistake is $e^{2x}$, i already edit
– LTY
Nov 26 '18 at 14:08












2 Answers
2






active

oldest

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1














The solution $y_1$ and $y_2$ are solutions to the homogeneous differential equation, that means, the right side is $0$. What you are confusing this is with a Particular solution. Refer to this on the solution of non-homogenous DE, http://www.math.psu.edu/tseng/class/Math251/Notes-2nd%20order%20ODE%20pt2.pdf






share|cite|improve this answer





















  • Does it means that the above question is about non-homogeneous DE? And the ways of solving it is different from hoemogeneous DE?
    – LTY
    Nov 26 '18 at 13:43



















0














This problem is about second-order linear equations. Usually, in textbooks, solving starts from the property
$y=y_c+y_p$,



where yc is a solution of homogeneous equation, and yp is a solution of nonhomogeneous equation. So $y_1$ and $y_2$ are not final solutions; they're rather two possible variants of $y_c$.



Now let's look at the problem from this point of view. By definition, if your nonhomogeneous de=f(x), then $y_c$ is solution for the situation when homogeneous de=0. Simply replace x^2*e^2x with 0.



Then, xy''-(1+x)y'+y=0 has solutions y1 and y2. Let's check.



$$x*e^x-(1+x)*e^x+e^x=0$$



$$x*0-(1+x)*1+(1+x)=0$$
Both are true.






share|cite|improve this answer























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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    The solution $y_1$ and $y_2$ are solutions to the homogeneous differential equation, that means, the right side is $0$. What you are confusing this is with a Particular solution. Refer to this on the solution of non-homogenous DE, http://www.math.psu.edu/tseng/class/Math251/Notes-2nd%20order%20ODE%20pt2.pdf






    share|cite|improve this answer





















    • Does it means that the above question is about non-homogeneous DE? And the ways of solving it is different from hoemogeneous DE?
      – LTY
      Nov 26 '18 at 13:43
















    1














    The solution $y_1$ and $y_2$ are solutions to the homogeneous differential equation, that means, the right side is $0$. What you are confusing this is with a Particular solution. Refer to this on the solution of non-homogenous DE, http://www.math.psu.edu/tseng/class/Math251/Notes-2nd%20order%20ODE%20pt2.pdf






    share|cite|improve this answer





















    • Does it means that the above question is about non-homogeneous DE? And the ways of solving it is different from hoemogeneous DE?
      – LTY
      Nov 26 '18 at 13:43














    1












    1








    1






    The solution $y_1$ and $y_2$ are solutions to the homogeneous differential equation, that means, the right side is $0$. What you are confusing this is with a Particular solution. Refer to this on the solution of non-homogenous DE, http://www.math.psu.edu/tseng/class/Math251/Notes-2nd%20order%20ODE%20pt2.pdf






    share|cite|improve this answer












    The solution $y_1$ and $y_2$ are solutions to the homogeneous differential equation, that means, the right side is $0$. What you are confusing this is with a Particular solution. Refer to this on the solution of non-homogenous DE, http://www.math.psu.edu/tseng/class/Math251/Notes-2nd%20order%20ODE%20pt2.pdf







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 26 '18 at 13:26









    Abhinav Jha

    193111




    193111












    • Does it means that the above question is about non-homogeneous DE? And the ways of solving it is different from hoemogeneous DE?
      – LTY
      Nov 26 '18 at 13:43


















    • Does it means that the above question is about non-homogeneous DE? And the ways of solving it is different from hoemogeneous DE?
      – LTY
      Nov 26 '18 at 13:43
















    Does it means that the above question is about non-homogeneous DE? And the ways of solving it is different from hoemogeneous DE?
    – LTY
    Nov 26 '18 at 13:43




    Does it means that the above question is about non-homogeneous DE? And the ways of solving it is different from hoemogeneous DE?
    – LTY
    Nov 26 '18 at 13:43











    0














    This problem is about second-order linear equations. Usually, in textbooks, solving starts from the property
    $y=y_c+y_p$,



    where yc is a solution of homogeneous equation, and yp is a solution of nonhomogeneous equation. So $y_1$ and $y_2$ are not final solutions; they're rather two possible variants of $y_c$.



    Now let's look at the problem from this point of view. By definition, if your nonhomogeneous de=f(x), then $y_c$ is solution for the situation when homogeneous de=0. Simply replace x^2*e^2x with 0.



    Then, xy''-(1+x)y'+y=0 has solutions y1 and y2. Let's check.



    $$x*e^x-(1+x)*e^x+e^x=0$$



    $$x*0-(1+x)*1+(1+x)=0$$
    Both are true.






    share|cite|improve this answer




























      0














      This problem is about second-order linear equations. Usually, in textbooks, solving starts from the property
      $y=y_c+y_p$,



      where yc is a solution of homogeneous equation, and yp is a solution of nonhomogeneous equation. So $y_1$ and $y_2$ are not final solutions; they're rather two possible variants of $y_c$.



      Now let's look at the problem from this point of view. By definition, if your nonhomogeneous de=f(x), then $y_c$ is solution for the situation when homogeneous de=0. Simply replace x^2*e^2x with 0.



      Then, xy''-(1+x)y'+y=0 has solutions y1 and y2. Let's check.



      $$x*e^x-(1+x)*e^x+e^x=0$$



      $$x*0-(1+x)*1+(1+x)=0$$
      Both are true.






      share|cite|improve this answer


























        0












        0








        0






        This problem is about second-order linear equations. Usually, in textbooks, solving starts from the property
        $y=y_c+y_p$,



        where yc is a solution of homogeneous equation, and yp is a solution of nonhomogeneous equation. So $y_1$ and $y_2$ are not final solutions; they're rather two possible variants of $y_c$.



        Now let's look at the problem from this point of view. By definition, if your nonhomogeneous de=f(x), then $y_c$ is solution for the situation when homogeneous de=0. Simply replace x^2*e^2x with 0.



        Then, xy''-(1+x)y'+y=0 has solutions y1 and y2. Let's check.



        $$x*e^x-(1+x)*e^x+e^x=0$$



        $$x*0-(1+x)*1+(1+x)=0$$
        Both are true.






        share|cite|improve this answer














        This problem is about second-order linear equations. Usually, in textbooks, solving starts from the property
        $y=y_c+y_p$,



        where yc is a solution of homogeneous equation, and yp is a solution of nonhomogeneous equation. So $y_1$ and $y_2$ are not final solutions; they're rather two possible variants of $y_c$.



        Now let's look at the problem from this point of view. By definition, if your nonhomogeneous de=f(x), then $y_c$ is solution for the situation when homogeneous de=0. Simply replace x^2*e^2x with 0.



        Then, xy''-(1+x)y'+y=0 has solutions y1 and y2. Let's check.



        $$x*e^x-(1+x)*e^x+e^x=0$$



        $$x*0-(1+x)*1+(1+x)=0$$
        Both are true.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 26 '18 at 17:36

























        answered Nov 26 '18 at 16:01









        Kelly Shepphard

        2298




        2298






























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