Solution of $y''-y'=xe^x$ using Method of Undetermined Coefficients












1














The problem gives $y''-y'=xe^x$ with conditions $y(0)=2$ and $y'left(0right)=1$.



My issue is when calculating the particular solution of the equation. I assumed that it should take the form $y_p=(Ax+B)e^x$. After substituting into the 2nd ODE, it becomes:



$(2A+Ax+B)e^x-(A+Ax+B)e^x=xe^x$



This simplifies to $Ae^x=xe^x$. I'm not sure how to progress. Is $A=0$? This does not seem like the correct way to proceed with solving the problem but I am not sure how to proceed.






differential-equations







share|cite|improve this question




















  • 3




    Try $(Ax^2+Bx+C)e^x$.
    – Lord Shark the Unknown
    Jun 28 '17 at 4:20










  • @LordSharktheUnknown Why wouldn't it take the form $(Ax+B)e^x$ though?
    – Matt Chu
    Jun 28 '17 at 4:21






  • 5




    I thought you'd just proved why not.
    – Lord Shark the Unknown
    Jun 28 '17 at 4:26










  • It doesn't take the form $(Ax + B)e^x because of the second derivative.
    – Retired account
    Jun 28 '17 at 4:43










  • The reason you don't have $(Ax+B)e^x$ is that $e^x$ is already a solution to the homogeneous equation $y''-y'=0$, so you need to find another term to use instead.
    – Carmeister
    Jun 28 '17 at 5:19
















1














The problem gives $y''-y'=xe^x$ with conditions $y(0)=2$ and $y'left(0right)=1$.



My issue is when calculating the particular solution of the equation. I assumed that it should take the form $y_p=(Ax+B)e^x$. After substituting into the 2nd ODE, it becomes:



$(2A+Ax+B)e^x-(A+Ax+B)e^x=xe^x$



This simplifies to $Ae^x=xe^x$. I'm not sure how to progress. Is $A=0$? This does not seem like the correct way to proceed with solving the problem but I am not sure how to proceed.






differential-equations







share|cite|improve this question




















  • 3




    Try $(Ax^2+Bx+C)e^x$.
    – Lord Shark the Unknown
    Jun 28 '17 at 4:20










  • @LordSharktheUnknown Why wouldn't it take the form $(Ax+B)e^x$ though?
    – Matt Chu
    Jun 28 '17 at 4:21






  • 5




    I thought you'd just proved why not.
    – Lord Shark the Unknown
    Jun 28 '17 at 4:26










  • It doesn't take the form $(Ax + B)e^x because of the second derivative.
    – Retired account
    Jun 28 '17 at 4:43










  • The reason you don't have $(Ax+B)e^x$ is that $e^x$ is already a solution to the homogeneous equation $y''-y'=0$, so you need to find another term to use instead.
    – Carmeister
    Jun 28 '17 at 5:19














1












1








1







The problem gives $y''-y'=xe^x$ with conditions $y(0)=2$ and $y'left(0right)=1$.



My issue is when calculating the particular solution of the equation. I assumed that it should take the form $y_p=(Ax+B)e^x$. After substituting into the 2nd ODE, it becomes:



$(2A+Ax+B)e^x-(A+Ax+B)e^x=xe^x$



This simplifies to $Ae^x=xe^x$. I'm not sure how to progress. Is $A=0$? This does not seem like the correct way to proceed with solving the problem but I am not sure how to proceed.






differential-equations







share|cite|improve this question















The problem gives $y''-y'=xe^x$ with conditions $y(0)=2$ and $y'left(0right)=1$.



My issue is when calculating the particular solution of the equation. I assumed that it should take the form $y_p=(Ax+B)e^x$. After substituting into the 2nd ODE, it becomes:



$(2A+Ax+B)e^x-(A+Ax+B)e^x=xe^x$



This simplifies to $Ae^x=xe^x$. I'm not sure how to progress. Is $A=0$? This does not seem like the correct way to proceed with solving the problem but I am not sure how to proceed.






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 Jun 28 '17 at 4:23

























asked Jun 28 '17 at 4:18









Matt Chu

112




112








  • 3




    Try $(Ax^2+Bx+C)e^x$.
    – Lord Shark the Unknown
    Jun 28 '17 at 4:20










  • @LordSharktheUnknown Why wouldn't it take the form $(Ax+B)e^x$ though?
    – Matt Chu
    Jun 28 '17 at 4:21






  • 5




    I thought you'd just proved why not.
    – Lord Shark the Unknown
    Jun 28 '17 at 4:26










  • It doesn't take the form $(Ax + B)e^x because of the second derivative.
    – Retired account
    Jun 28 '17 at 4:43










  • The reason you don't have $(Ax+B)e^x$ is that $e^x$ is already a solution to the homogeneous equation $y''-y'=0$, so you need to find another term to use instead.
    – Carmeister
    Jun 28 '17 at 5:19














  • 3




    Try $(Ax^2+Bx+C)e^x$.
    – Lord Shark the Unknown
    Jun 28 '17 at 4:20










  • @LordSharktheUnknown Why wouldn't it take the form $(Ax+B)e^x$ though?
    – Matt Chu
    Jun 28 '17 at 4:21






  • 5




    I thought you'd just proved why not.
    – Lord Shark the Unknown
    Jun 28 '17 at 4:26










  • It doesn't take the form $(Ax + B)e^x because of the second derivative.
    – Retired account
    Jun 28 '17 at 4:43










  • The reason you don't have $(Ax+B)e^x$ is that $e^x$ is already a solution to the homogeneous equation $y''-y'=0$, so you need to find another term to use instead.
    – Carmeister
    Jun 28 '17 at 5:19








3




3




Try $(Ax^2+Bx+C)e^x$.
– Lord Shark the Unknown
Jun 28 '17 at 4:20




Try $(Ax^2+Bx+C)e^x$.
– Lord Shark the Unknown
Jun 28 '17 at 4:20












@LordSharktheUnknown Why wouldn't it take the form $(Ax+B)e^x$ though?
– Matt Chu
Jun 28 '17 at 4:21




@LordSharktheUnknown Why wouldn't it take the form $(Ax+B)e^x$ though?
– Matt Chu
Jun 28 '17 at 4:21




5




5




I thought you'd just proved why not.
– Lord Shark the Unknown
Jun 28 '17 at 4:26




I thought you'd just proved why not.
– Lord Shark the Unknown
Jun 28 '17 at 4:26












It doesn't take the form $(Ax + B)e^x because of the second derivative.
– Retired account
Jun 28 '17 at 4:43




It doesn't take the form $(Ax + B)e^x because of the second derivative.
– Retired account
Jun 28 '17 at 4:43












The reason you don't have $(Ax+B)e^x$ is that $e^x$ is already a solution to the homogeneous equation $y''-y'=0$, so you need to find another term to use instead.
– Carmeister
Jun 28 '17 at 5:19




The reason you don't have $(Ax+B)e^x$ is that $e^x$ is already a solution to the homogeneous equation $y''-y'=0$, so you need to find another term to use instead.
– Carmeister
Jun 28 '17 at 5:19










2 Answers
2






active

oldest

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1














If I may suggest, proceed by step.



Start using $y'=u$ to have $$y''-y'=xe^ximplies u'-u=xe^x$$ Now let $u=v e^x$ to get $$v' e^x=x e^x implies v'=ximplies v=frac 12 x^2+c_1$$ Here, you already see that the solution will contain an $x^2$ term somewhere.






share|cite|improve this answer





























    0














    For any Problem First try taking the RHS as y, In this case y=xe^x. Substitute in the LHS, If you get the answer as 0 then your guess is simply not compatible, It should be yp=(Ax+B)xe^x. To be exact you should multiply your guess by x until you find a different league of terms than in the LHS.






    share|cite|improve this answer





















    • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
      – José Carlos Santos
      Nov 28 '18 at 8:58











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    If I may suggest, proceed by step.



    Start using $y'=u$ to have $$y''-y'=xe^ximplies u'-u=xe^x$$ Now let $u=v e^x$ to get $$v' e^x=x e^x implies v'=ximplies v=frac 12 x^2+c_1$$ Here, you already see that the solution will contain an $x^2$ term somewhere.






    share|cite|improve this answer


























      1














      If I may suggest, proceed by step.



      Start using $y'=u$ to have $$y''-y'=xe^ximplies u'-u=xe^x$$ Now let $u=v e^x$ to get $$v' e^x=x e^x implies v'=ximplies v=frac 12 x^2+c_1$$ Here, you already see that the solution will contain an $x^2$ term somewhere.






      share|cite|improve this answer
























        1












        1








        1






        If I may suggest, proceed by step.



        Start using $y'=u$ to have $$y''-y'=xe^ximplies u'-u=xe^x$$ Now let $u=v e^x$ to get $$v' e^x=x e^x implies v'=ximplies v=frac 12 x^2+c_1$$ Here, you already see that the solution will contain an $x^2$ term somewhere.






        share|cite|improve this answer












        If I may suggest, proceed by step.



        Start using $y'=u$ to have $$y''-y'=xe^ximplies u'-u=xe^x$$ Now let $u=v e^x$ to get $$v' e^x=x e^x implies v'=ximplies v=frac 12 x^2+c_1$$ Here, you already see that the solution will contain an $x^2$ term somewhere.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jun 28 '17 at 4:59









        Claude Leibovici

        119k1157132




        119k1157132























            0














            For any Problem First try taking the RHS as y, In this case y=xe^x. Substitute in the LHS, If you get the answer as 0 then your guess is simply not compatible, It should be yp=(Ax+B)xe^x. To be exact you should multiply your guess by x until you find a different league of terms than in the LHS.






            share|cite|improve this answer





















            • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
              – José Carlos Santos
              Nov 28 '18 at 8:58
















            0














            For any Problem First try taking the RHS as y, In this case y=xe^x. Substitute in the LHS, If you get the answer as 0 then your guess is simply not compatible, It should be yp=(Ax+B)xe^x. To be exact you should multiply your guess by x until you find a different league of terms than in the LHS.






            share|cite|improve this answer





















            • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
              – José Carlos Santos
              Nov 28 '18 at 8:58














            0












            0








            0






            For any Problem First try taking the RHS as y, In this case y=xe^x. Substitute in the LHS, If you get the answer as 0 then your guess is simply not compatible, It should be yp=(Ax+B)xe^x. To be exact you should multiply your guess by x until you find a different league of terms than in the LHS.






            share|cite|improve this answer












            For any Problem First try taking the RHS as y, In this case y=xe^x. Substitute in the LHS, If you get the answer as 0 then your guess is simply not compatible, It should be yp=(Ax+B)xe^x. To be exact you should multiply your guess by x until you find a different league of terms than in the LHS.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 28 '18 at 8:37









            AwWSaM

            1




            1












            • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
              – José Carlos Santos
              Nov 28 '18 at 8:58


















            • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
              – José Carlos Santos
              Nov 28 '18 at 8:58
















            Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
            – José Carlos Santos
            Nov 28 '18 at 8:58




            Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
            – José Carlos Santos
            Nov 28 '18 at 8:58


















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