solving Second Order Non-homogeneous Cauchy-Euler Differential Equations












0












$begingroup$


$2x^2y''-11xy'-7y=3x^7$



It has two real solutions : $7,-frac{1}{2}$



I got the general solution for the homogeneous part: $y_g =c_1x^7+frac{c_2}{sqrt{x}}$



Now I want to get the particular solution for $y_g=3x^7$



Which method is best here? I used The "variation of parameters method" but I think I got stuck with the integral:



$W(x^7,x^{frac{1}{2}})=...=-frac{1}{2}x^{frac{15}{2}}-7x^{frac{11}{2}}$



$y_p= -x^7intfrac{x^{-frac{1}{2}}3x^5}{-frac{1}{2}x^{frac{15}{2}}-7x^{frac{11}{2}}}dx+x^{-frac{1}{2}}intfrac{x^73x^5}{-frac{1}{2}x^{frac{15}{2}}-7x^{frac{11}{2}}}dx$



Am I doing this right?






ordinary-differential-equations







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  • 1




    $begingroup$
    You can use undertermined coefficients here. The particular solution has the form $y_p(x)= Ax^7ln x$
    $endgroup$
    – Dylan
    Jan 19 at 15:27












  • $begingroup$
    @Dylan, Yup It works! That's even simpler.Thank you
    $endgroup$
    – NPLS
    Jan 19 at 16:13
















0












$begingroup$


$2x^2y''-11xy'-7y=3x^7$



It has two real solutions : $7,-frac{1}{2}$



I got the general solution for the homogeneous part: $y_g =c_1x^7+frac{c_2}{sqrt{x}}$



Now I want to get the particular solution for $y_g=3x^7$



Which method is best here? I used The "variation of parameters method" but I think I got stuck with the integral:



$W(x^7,x^{frac{1}{2}})=...=-frac{1}{2}x^{frac{15}{2}}-7x^{frac{11}{2}}$



$y_p= -x^7intfrac{x^{-frac{1}{2}}3x^5}{-frac{1}{2}x^{frac{15}{2}}-7x^{frac{11}{2}}}dx+x^{-frac{1}{2}}intfrac{x^73x^5}{-frac{1}{2}x^{frac{15}{2}}-7x^{frac{11}{2}}}dx$



Am I doing this right?






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You can use undertermined coefficients here. The particular solution has the form $y_p(x)= Ax^7ln x$
    $endgroup$
    – Dylan
    Jan 19 at 15:27












  • $begingroup$
    @Dylan, Yup It works! That's even simpler.Thank you
    $endgroup$
    – NPLS
    Jan 19 at 16:13














0












0








0





$begingroup$


$2x^2y''-11xy'-7y=3x^7$



It has two real solutions : $7,-frac{1}{2}$



I got the general solution for the homogeneous part: $y_g =c_1x^7+frac{c_2}{sqrt{x}}$



Now I want to get the particular solution for $y_g=3x^7$



Which method is best here? I used The "variation of parameters method" but I think I got stuck with the integral:



$W(x^7,x^{frac{1}{2}})=...=-frac{1}{2}x^{frac{15}{2}}-7x^{frac{11}{2}}$



$y_p= -x^7intfrac{x^{-frac{1}{2}}3x^5}{-frac{1}{2}x^{frac{15}{2}}-7x^{frac{11}{2}}}dx+x^{-frac{1}{2}}intfrac{x^73x^5}{-frac{1}{2}x^{frac{15}{2}}-7x^{frac{11}{2}}}dx$



Am I doing this right?






ordinary-differential-equations







share|cite|improve this question









$endgroup$




$2x^2y''-11xy'-7y=3x^7$



It has two real solutions : $7,-frac{1}{2}$



I got the general solution for the homogeneous part: $y_g =c_1x^7+frac{c_2}{sqrt{x}}$



Now I want to get the particular solution for $y_g=3x^7$



Which method is best here? I used The "variation of parameters method" but I think I got stuck with the integral:



$W(x^7,x^{frac{1}{2}})=...=-frac{1}{2}x^{frac{15}{2}}-7x^{frac{11}{2}}$



$y_p= -x^7intfrac{x^{-frac{1}{2}}3x^5}{-frac{1}{2}x^{frac{15}{2}}-7x^{frac{11}{2}}}dx+x^{-frac{1}{2}}intfrac{x^73x^5}{-frac{1}{2}x^{frac{15}{2}}-7x^{frac{11}{2}}}dx$



Am I doing this right?






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 19 at 14:44









NPLSNPLS

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7112








  • 1




    $begingroup$
    You can use undertermined coefficients here. The particular solution has the form $y_p(x)= Ax^7ln x$
    $endgroup$
    – Dylan
    Jan 19 at 15:27












  • $begingroup$
    @Dylan, Yup It works! That's even simpler.Thank you
    $endgroup$
    – NPLS
    Jan 19 at 16:13














  • 1




    $begingroup$
    You can use undertermined coefficients here. The particular solution has the form $y_p(x)= Ax^7ln x$
    $endgroup$
    – Dylan
    Jan 19 at 15:27












  • $begingroup$
    @Dylan, Yup It works! That's even simpler.Thank you
    $endgroup$
    – NPLS
    Jan 19 at 16:13








1




1




$begingroup$
You can use undertermined coefficients here. The particular solution has the form $y_p(x)= Ax^7ln x$
$endgroup$
– Dylan
Jan 19 at 15:27






$begingroup$
You can use undertermined coefficients here. The particular solution has the form $y_p(x)= Ax^7ln x$
$endgroup$
– Dylan
Jan 19 at 15:27














$begingroup$
@Dylan, Yup It works! That's even simpler.Thank you
$endgroup$
– NPLS
Jan 19 at 16:13




$begingroup$
@Dylan, Yup It works! That's even simpler.Thank you
$endgroup$
– NPLS
Jan 19 at 16:13










1 Answer
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$begingroup$

Make the ansatz $$y(x)=x^r$$ then you have to solve $$2r^2-13r-7=0$$ and you get
$$y(x)=frac{C_1}{sqrt{x}}+C_2x^7$$ a particular solution can be find by the method of variation of the constants. It is $$y_p=-frac{2x^7}{75}+frac{1}{5}x^7log(x)$$
Let $$y_{b_1}=frac{1}{sqrt{x}},y_{b_2}=x^7$$ then we get
$$W(x)=begin{vmatrix}frac{1}{sqrt{x}}&x^7\frac{d}{dx}(frac{1}{sqrt{x}})&frac{d}{dx}(x^7)end{vmatrix}$$
And this is $$frac{15}{2}x^{frac{11}{2}}$$. Let $$f(x)=frac{3}{2}x^5$$ then we obtain
$$v_1=-intfrac{f(x)y_{b_2}}{W(x)}dx$$ and $$v_2=frac{f(x)y_{b_1}(x)}{W(x)}dx$$
and we get
$$y_p(x)=v_1y_{b_1}(x)+v_2(x)y_{b_1}(x)$$ this gives the result above.






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    $begingroup$

    Make the ansatz $$y(x)=x^r$$ then you have to solve $$2r^2-13r-7=0$$ and you get
    $$y(x)=frac{C_1}{sqrt{x}}+C_2x^7$$ a particular solution can be find by the method of variation of the constants. It is $$y_p=-frac{2x^7}{75}+frac{1}{5}x^7log(x)$$
    Let $$y_{b_1}=frac{1}{sqrt{x}},y_{b_2}=x^7$$ then we get
    $$W(x)=begin{vmatrix}frac{1}{sqrt{x}}&x^7\frac{d}{dx}(frac{1}{sqrt{x}})&frac{d}{dx}(x^7)end{vmatrix}$$
    And this is $$frac{15}{2}x^{frac{11}{2}}$$. Let $$f(x)=frac{3}{2}x^5$$ then we obtain
    $$v_1=-intfrac{f(x)y_{b_2}}{W(x)}dx$$ and $$v_2=frac{f(x)y_{b_1}(x)}{W(x)}dx$$
    and we get
    $$y_p(x)=v_1y_{b_1}(x)+v_2(x)y_{b_1}(x)$$ this gives the result above.






    share|cite|improve this answer











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      0












      $begingroup$

      Make the ansatz $$y(x)=x^r$$ then you have to solve $$2r^2-13r-7=0$$ and you get
      $$y(x)=frac{C_1}{sqrt{x}}+C_2x^7$$ a particular solution can be find by the method of variation of the constants. It is $$y_p=-frac{2x^7}{75}+frac{1}{5}x^7log(x)$$
      Let $$y_{b_1}=frac{1}{sqrt{x}},y_{b_2}=x^7$$ then we get
      $$W(x)=begin{vmatrix}frac{1}{sqrt{x}}&x^7\frac{d}{dx}(frac{1}{sqrt{x}})&frac{d}{dx}(x^7)end{vmatrix}$$
      And this is $$frac{15}{2}x^{frac{11}{2}}$$. Let $$f(x)=frac{3}{2}x^5$$ then we obtain
      $$v_1=-intfrac{f(x)y_{b_2}}{W(x)}dx$$ and $$v_2=frac{f(x)y_{b_1}(x)}{W(x)}dx$$
      and we get
      $$y_p(x)=v_1y_{b_1}(x)+v_2(x)y_{b_1}(x)$$ this gives the result above.






      share|cite|improve this answer











      $endgroup$
















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        0





        $begingroup$

        Make the ansatz $$y(x)=x^r$$ then you have to solve $$2r^2-13r-7=0$$ and you get
        $$y(x)=frac{C_1}{sqrt{x}}+C_2x^7$$ a particular solution can be find by the method of variation of the constants. It is $$y_p=-frac{2x^7}{75}+frac{1}{5}x^7log(x)$$
        Let $$y_{b_1}=frac{1}{sqrt{x}},y_{b_2}=x^7$$ then we get
        $$W(x)=begin{vmatrix}frac{1}{sqrt{x}}&x^7\frac{d}{dx}(frac{1}{sqrt{x}})&frac{d}{dx}(x^7)end{vmatrix}$$
        And this is $$frac{15}{2}x^{frac{11}{2}}$$. Let $$f(x)=frac{3}{2}x^5$$ then we obtain
        $$v_1=-intfrac{f(x)y_{b_2}}{W(x)}dx$$ and $$v_2=frac{f(x)y_{b_1}(x)}{W(x)}dx$$
        and we get
        $$y_p(x)=v_1y_{b_1}(x)+v_2(x)y_{b_1}(x)$$ this gives the result above.






        share|cite|improve this answer











        $endgroup$



        Make the ansatz $$y(x)=x^r$$ then you have to solve $$2r^2-13r-7=0$$ and you get
        $$y(x)=frac{C_1}{sqrt{x}}+C_2x^7$$ a particular solution can be find by the method of variation of the constants. It is $$y_p=-frac{2x^7}{75}+frac{1}{5}x^7log(x)$$
        Let $$y_{b_1}=frac{1}{sqrt{x}},y_{b_2}=x^7$$ then we get
        $$W(x)=begin{vmatrix}frac{1}{sqrt{x}}&x^7\frac{d}{dx}(frac{1}{sqrt{x}})&frac{d}{dx}(x^7)end{vmatrix}$$
        And this is $$frac{15}{2}x^{frac{11}{2}}$$. Let $$f(x)=frac{3}{2}x^5$$ then we obtain
        $$v_1=-intfrac{f(x)y_{b_2}}{W(x)}dx$$ and $$v_2=frac{f(x)y_{b_1}(x)}{W(x)}dx$$
        and we get
        $$y_p(x)=v_1y_{b_1}(x)+v_2(x)y_{b_1}(x)$$ this gives the result above.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 19 at 15:27

























        answered Jan 19 at 14:53









        Dr. Sonnhard GraubnerDr. Sonnhard Graubner

        75.4k42866




        75.4k42866






























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