Solution to Differential equation 2nd order, x*e^x












2












$begingroup$


I'm trying to solve the following differential equation:



$y'' + y = x cdot e^x$



I already have the homogeneous solution: $y_h = c_1 cdot cos(x) + c_2 cdot sin(x)$ but I'm struggeling to find the particular solution.
I tried using $y_p = a cdot x cdot e^x$ but that didn't quite work, or maybe I did something wrong...



Can somebody explain to me how to solve this?



Thanks.



Edit: Here's the solution:



$y_p = (a cdot x + b) cdot e^x$



$y_p' = (a + a cdot x + b) cdot e^x$ and $ y_p'' = (2 cdot a + a cdot x + b) cdot e^x$



$y_p'' + y_p = (2 cdot a cdot x + 2 cdot a + 2 cdot b) cdot e^x$



$Rightarrow 2 cdot a = 1 Rightarrow a = frac{1}{2}$ and $2 cdot a + 2 cdot b = 0 Leftrightarrow 2 cdot a = 1 = - 2 cdot b Rightarrow b = - frac{1}{2}$



$Rightarrow y_p = (a cdot x + b) cdot e^x = frac{x-1}{2} cdot e^x$



All in all: $y = y_h + y_p = c_1 cdot cos(x) + c_2 cdot sin(x) + frac{x-1}{2} cdot e^x$






ordinary-differential-equations exponential-function







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I'm trying to solve the following differential equation:



    $y'' + y = x cdot e^x$



    I already have the homogeneous solution: $y_h = c_1 cdot cos(x) + c_2 cdot sin(x)$ but I'm struggeling to find the particular solution.
    I tried using $y_p = a cdot x cdot e^x$ but that didn't quite work, or maybe I did something wrong...



    Can somebody explain to me how to solve this?



    Thanks.



    Edit: Here's the solution:



    $y_p = (a cdot x + b) cdot e^x$



    $y_p' = (a + a cdot x + b) cdot e^x$ and $ y_p'' = (2 cdot a + a cdot x + b) cdot e^x$



    $y_p'' + y_p = (2 cdot a cdot x + 2 cdot a + 2 cdot b) cdot e^x$



    $Rightarrow 2 cdot a = 1 Rightarrow a = frac{1}{2}$ and $2 cdot a + 2 cdot b = 0 Leftrightarrow 2 cdot a = 1 = - 2 cdot b Rightarrow b = - frac{1}{2}$



    $Rightarrow y_p = (a cdot x + b) cdot e^x = frac{x-1}{2} cdot e^x$



    All in all: $y = y_h + y_p = c_1 cdot cos(x) + c_2 cdot sin(x) + frac{x-1}{2} cdot e^x$






    ordinary-differential-equations exponential-function







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I'm trying to solve the following differential equation:



      $y'' + y = x cdot e^x$



      I already have the homogeneous solution: $y_h = c_1 cdot cos(x) + c_2 cdot sin(x)$ but I'm struggeling to find the particular solution.
      I tried using $y_p = a cdot x cdot e^x$ but that didn't quite work, or maybe I did something wrong...



      Can somebody explain to me how to solve this?



      Thanks.



      Edit: Here's the solution:



      $y_p = (a cdot x + b) cdot e^x$



      $y_p' = (a + a cdot x + b) cdot e^x$ and $ y_p'' = (2 cdot a + a cdot x + b) cdot e^x$



      $y_p'' + y_p = (2 cdot a cdot x + 2 cdot a + 2 cdot b) cdot e^x$



      $Rightarrow 2 cdot a = 1 Rightarrow a = frac{1}{2}$ and $2 cdot a + 2 cdot b = 0 Leftrightarrow 2 cdot a = 1 = - 2 cdot b Rightarrow b = - frac{1}{2}$



      $Rightarrow y_p = (a cdot x + b) cdot e^x = frac{x-1}{2} cdot e^x$



      All in all: $y = y_h + y_p = c_1 cdot cos(x) + c_2 cdot sin(x) + frac{x-1}{2} cdot e^x$






      ordinary-differential-equations exponential-function







      share|cite|improve this question











      $endgroup$




      I'm trying to solve the following differential equation:



      $y'' + y = x cdot e^x$



      I already have the homogeneous solution: $y_h = c_1 cdot cos(x) + c_2 cdot sin(x)$ but I'm struggeling to find the particular solution.
      I tried using $y_p = a cdot x cdot e^x$ but that didn't quite work, or maybe I did something wrong...



      Can somebody explain to me how to solve this?



      Thanks.



      Edit: Here's the solution:



      $y_p = (a cdot x + b) cdot e^x$



      $y_p' = (a + a cdot x + b) cdot e^x$ and $ y_p'' = (2 cdot a + a cdot x + b) cdot e^x$



      $y_p'' + y_p = (2 cdot a cdot x + 2 cdot a + 2 cdot b) cdot e^x$



      $Rightarrow 2 cdot a = 1 Rightarrow a = frac{1}{2}$ and $2 cdot a + 2 cdot b = 0 Leftrightarrow 2 cdot a = 1 = - 2 cdot b Rightarrow b = - frac{1}{2}$



      $Rightarrow y_p = (a cdot x + b) cdot e^x = frac{x-1}{2} cdot e^x$



      All in all: $y = y_h + y_p = c_1 cdot cos(x) + c_2 cdot sin(x) + frac{x-1}{2} cdot e^x$






      ordinary-differential-equations exponential-function




      ordinary-differential-equations exponential-function






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 21 at 17:11







      zutru

















      asked Jan 21 at 15:09









      zutruzutru

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          $begingroup$

          Try $y_p = (ax+b)e^x$.



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            $begingroup$
            oh.. yeah, that works, thanks
            $endgroup$
            – zutru
            Jan 21 at 17:11











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          $begingroup$

          Try $y_p = (ax+b)e^x$.



          [Additional characters to bypass the 30 characters rule]






          share|cite|improve this answer









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            $begingroup$
            oh.. yeah, that works, thanks
            $endgroup$
            – zutru
            Jan 21 at 17:11
















          1












          $begingroup$

          Try $y_p = (ax+b)e^x$.



          [Additional characters to bypass the 30 characters rule]






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            oh.. yeah, that works, thanks
            $endgroup$
            – zutru
            Jan 21 at 17:11














          1












          1








          1





          $begingroup$

          Try $y_p = (ax+b)e^x$.



          [Additional characters to bypass the 30 characters rule]






          share|cite|improve this answer









          $endgroup$



          Try $y_p = (ax+b)e^x$.



          [Additional characters to bypass the 30 characters rule]







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 21 at 15:12









          KlausKlaus

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          2,12711








          • 1




            $begingroup$
            oh.. yeah, that works, thanks
            $endgroup$
            – zutru
            Jan 21 at 17:11














          • 1




            $begingroup$
            oh.. yeah, that works, thanks
            $endgroup$
            – zutru
            Jan 21 at 17:11








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          1




          $begingroup$
          oh.. yeah, that works, thanks
          $endgroup$
          – zutru
          Jan 21 at 17:11




          $begingroup$
          oh.. yeah, that works, thanks
          $endgroup$
          – zutru
          Jan 21 at 17:11


















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