Find the general solution to the equation $y^{''}-6y^{'}+8y=5xe^{2x}+2e^{4x}sin x$












0












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Solving the homogeneous equation
$$y^{''}-6y^{'}+8=0Rightarrow lambda^2-6lambda+8=0Rightarrow lambda_1=2,lambda_2=4$$



General solution to this homogeneous equation is $y_h=c_1e^{2x}+c_2e^{4x}$.



How to find one particular solution to the non-homogeneous equation?






calculus ordinary-differential-equations







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    0












    $begingroup$


    Solving the homogeneous equation
    $$y^{''}-6y^{'}+8=0Rightarrow lambda^2-6lambda+8=0Rightarrow lambda_1=2,lambda_2=4$$



    General solution to this homogeneous equation is $y_h=c_1e^{2x}+c_2e^{4x}$.



    How to find one particular solution to the non-homogeneous equation?






    calculus ordinary-differential-equations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Solving the homogeneous equation
      $$y^{''}-6y^{'}+8=0Rightarrow lambda^2-6lambda+8=0Rightarrow lambda_1=2,lambda_2=4$$



      General solution to this homogeneous equation is $y_h=c_1e^{2x}+c_2e^{4x}$.



      How to find one particular solution to the non-homogeneous equation?






      calculus ordinary-differential-equations







      share|cite|improve this question









      $endgroup$




      Solving the homogeneous equation
      $$y^{''}-6y^{'}+8=0Rightarrow lambda^2-6lambda+8=0Rightarrow lambda_1=2,lambda_2=4$$



      General solution to this homogeneous equation is $y_h=c_1e^{2x}+c_2e^{4x}$.



      How to find one particular solution to the non-homogeneous equation?






      calculus ordinary-differential-equations




      calculus ordinary-differential-equations






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 19 '16 at 22:36









      user300048user300048

      618412




      618412






















          2 Answers
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          $begingroup$

          With superposition of solutions:



          Find a particular solution $y_0(x)$ for $ y''-6y'+8=xmathrm e^{2x} $, a particular solution $ z_0(x) $ for $y''-6y'+8=xmathrm e^{4x}sin x$. A particular solution for $y''-6y'+8=5xmathrm e^{2x}+2mathrm e^{4x}sin x$ will be $5y_0(x)+2z_0(x)$.



          For a right-hand side of the form $p(x)mathrm e^{2x}$, $p(x)$ a polynomial of degree $d$, as $mathrm e^{2x}$ is a solution of the homogeneous equation, $y_0(x)$ will have the form $xq(x)mathrm e^{2x}$, with $deg q=d$.



          As for the right-hand side $mathrm e^{4x}sin x$, it is the imaginary part of $mathrm e^{(4+mathrm i)x}$, which leads to a particular solution of the form $Kmathrm e^{(4+mathrm i)x}$. Then take the imaginary part of this particular solution.






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            1












            $begingroup$

            Particular solution we find with method of undermined coefficients
            $$y_p={{e}^{4 x}}, left( A sin{(x)}+B cos{(x)}right) +{{e}^{2 x}} x, left( C x+Dright) $$
            We get
            $${{e}^{4 x}}, left( -2 B sin{(x)}-A sin{(x)}-B cos{(x)}+2 A cos{(x)}right) +{{e}^{2 x}}, left( -4 C x-2 D+2 Cright)\ =2 {{e}^{4 x}} sin{(x)}+5 {{e}^{2 x}} x,$$
            $$-2 B-A=2,quad 2 A-B=0,quad -4 C=5,quad 2 C-2 D=0,$$
            $$A=-frac{2}{5},quad B=-frac{4}{5},quad C=-frac{5}{4},quad D=-frac{5}{4},$$
            $$y_p={{e}^{4 x}}, left( -frac{2 sin{(x)}}{5}-frac{4 cos{(x)}}{5}right) +{{e}^{2 x}}, left( -frac{5 x}{4}-frac{5}{4}right) x\=
            -frac{2 {{e}^{4 x}}, left( sin{(x)}+2 cos{(x)}right) }{5}-frac{5 {{e}^{2 x}} x, left( x+1right) }{4},$$
            General solution is
            $$y=c_1e^{2x}+c_2e^{4x}+y_p$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              +1 for the detailed calculations. However, you should not "undermine" the "undetermined" coefficients.
              $endgroup$
              – farruhota
              Jan 27 at 10:03











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            2 Answers
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            2 Answers
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            active

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            active

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            2












            $begingroup$

            With superposition of solutions:



            Find a particular solution $y_0(x)$ for $ y''-6y'+8=xmathrm e^{2x} $, a particular solution $ z_0(x) $ for $y''-6y'+8=xmathrm e^{4x}sin x$. A particular solution for $y''-6y'+8=5xmathrm e^{2x}+2mathrm e^{4x}sin x$ will be $5y_0(x)+2z_0(x)$.



            For a right-hand side of the form $p(x)mathrm e^{2x}$, $p(x)$ a polynomial of degree $d$, as $mathrm e^{2x}$ is a solution of the homogeneous equation, $y_0(x)$ will have the form $xq(x)mathrm e^{2x}$, with $deg q=d$.



            As for the right-hand side $mathrm e^{4x}sin x$, it is the imaginary part of $mathrm e^{(4+mathrm i)x}$, which leads to a particular solution of the form $Kmathrm e^{(4+mathrm i)x}$. Then take the imaginary part of this particular solution.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              With superposition of solutions:



              Find a particular solution $y_0(x)$ for $ y''-6y'+8=xmathrm e^{2x} $, a particular solution $ z_0(x) $ for $y''-6y'+8=xmathrm e^{4x}sin x$. A particular solution for $y''-6y'+8=5xmathrm e^{2x}+2mathrm e^{4x}sin x$ will be $5y_0(x)+2z_0(x)$.



              For a right-hand side of the form $p(x)mathrm e^{2x}$, $p(x)$ a polynomial of degree $d$, as $mathrm e^{2x}$ is a solution of the homogeneous equation, $y_0(x)$ will have the form $xq(x)mathrm e^{2x}$, with $deg q=d$.



              As for the right-hand side $mathrm e^{4x}sin x$, it is the imaginary part of $mathrm e^{(4+mathrm i)x}$, which leads to a particular solution of the form $Kmathrm e^{(4+mathrm i)x}$. Then take the imaginary part of this particular solution.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                With superposition of solutions:



                Find a particular solution $y_0(x)$ for $ y''-6y'+8=xmathrm e^{2x} $, a particular solution $ z_0(x) $ for $y''-6y'+8=xmathrm e^{4x}sin x$. A particular solution for $y''-6y'+8=5xmathrm e^{2x}+2mathrm e^{4x}sin x$ will be $5y_0(x)+2z_0(x)$.



                For a right-hand side of the form $p(x)mathrm e^{2x}$, $p(x)$ a polynomial of degree $d$, as $mathrm e^{2x}$ is a solution of the homogeneous equation, $y_0(x)$ will have the form $xq(x)mathrm e^{2x}$, with $deg q=d$.



                As for the right-hand side $mathrm e^{4x}sin x$, it is the imaginary part of $mathrm e^{(4+mathrm i)x}$, which leads to a particular solution of the form $Kmathrm e^{(4+mathrm i)x}$. Then take the imaginary part of this particular solution.






                share|cite|improve this answer









                $endgroup$



                With superposition of solutions:



                Find a particular solution $y_0(x)$ for $ y''-6y'+8=xmathrm e^{2x} $, a particular solution $ z_0(x) $ for $y''-6y'+8=xmathrm e^{4x}sin x$. A particular solution for $y''-6y'+8=5xmathrm e^{2x}+2mathrm e^{4x}sin x$ will be $5y_0(x)+2z_0(x)$.



                For a right-hand side of the form $p(x)mathrm e^{2x}$, $p(x)$ a polynomial of degree $d$, as $mathrm e^{2x}$ is a solution of the homogeneous equation, $y_0(x)$ will have the form $xq(x)mathrm e^{2x}$, with $deg q=d$.



                As for the right-hand side $mathrm e^{4x}sin x$, it is the imaginary part of $mathrm e^{(4+mathrm i)x}$, which leads to a particular solution of the form $Kmathrm e^{(4+mathrm i)x}$. Then take the imaginary part of this particular solution.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 19 '16 at 22:51









                BernardBernard

                123k741117




                123k741117























                    1












                    $begingroup$

                    Particular solution we find with method of undermined coefficients
                    $$y_p={{e}^{4 x}}, left( A sin{(x)}+B cos{(x)}right) +{{e}^{2 x}} x, left( C x+Dright) $$
                    We get
                    $${{e}^{4 x}}, left( -2 B sin{(x)}-A sin{(x)}-B cos{(x)}+2 A cos{(x)}right) +{{e}^{2 x}}, left( -4 C x-2 D+2 Cright)\ =2 {{e}^{4 x}} sin{(x)}+5 {{e}^{2 x}} x,$$
                    $$-2 B-A=2,quad 2 A-B=0,quad -4 C=5,quad 2 C-2 D=0,$$
                    $$A=-frac{2}{5},quad B=-frac{4}{5},quad C=-frac{5}{4},quad D=-frac{5}{4},$$
                    $$y_p={{e}^{4 x}}, left( -frac{2 sin{(x)}}{5}-frac{4 cos{(x)}}{5}right) +{{e}^{2 x}}, left( -frac{5 x}{4}-frac{5}{4}right) x\=
                    -frac{2 {{e}^{4 x}}, left( sin{(x)}+2 cos{(x)}right) }{5}-frac{5 {{e}^{2 x}} x, left( x+1right) }{4},$$
                    General solution is
                    $$y=c_1e^{2x}+c_2e^{4x}+y_p$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      +1 for the detailed calculations. However, you should not "undermine" the "undetermined" coefficients.
                      $endgroup$
                      – farruhota
                      Jan 27 at 10:03
















                    1












                    $begingroup$

                    Particular solution we find with method of undermined coefficients
                    $$y_p={{e}^{4 x}}, left( A sin{(x)}+B cos{(x)}right) +{{e}^{2 x}} x, left( C x+Dright) $$
                    We get
                    $${{e}^{4 x}}, left( -2 B sin{(x)}-A sin{(x)}-B cos{(x)}+2 A cos{(x)}right) +{{e}^{2 x}}, left( -4 C x-2 D+2 Cright)\ =2 {{e}^{4 x}} sin{(x)}+5 {{e}^{2 x}} x,$$
                    $$-2 B-A=2,quad 2 A-B=0,quad -4 C=5,quad 2 C-2 D=0,$$
                    $$A=-frac{2}{5},quad B=-frac{4}{5},quad C=-frac{5}{4},quad D=-frac{5}{4},$$
                    $$y_p={{e}^{4 x}}, left( -frac{2 sin{(x)}}{5}-frac{4 cos{(x)}}{5}right) +{{e}^{2 x}}, left( -frac{5 x}{4}-frac{5}{4}right) x\=
                    -frac{2 {{e}^{4 x}}, left( sin{(x)}+2 cos{(x)}right) }{5}-frac{5 {{e}^{2 x}} x, left( x+1right) }{4},$$
                    General solution is
                    $$y=c_1e^{2x}+c_2e^{4x}+y_p$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      +1 for the detailed calculations. However, you should not "undermine" the "undetermined" coefficients.
                      $endgroup$
                      – farruhota
                      Jan 27 at 10:03














                    1












                    1








                    1





                    $begingroup$

                    Particular solution we find with method of undermined coefficients
                    $$y_p={{e}^{4 x}}, left( A sin{(x)}+B cos{(x)}right) +{{e}^{2 x}} x, left( C x+Dright) $$
                    We get
                    $${{e}^{4 x}}, left( -2 B sin{(x)}-A sin{(x)}-B cos{(x)}+2 A cos{(x)}right) +{{e}^{2 x}}, left( -4 C x-2 D+2 Cright)\ =2 {{e}^{4 x}} sin{(x)}+5 {{e}^{2 x}} x,$$
                    $$-2 B-A=2,quad 2 A-B=0,quad -4 C=5,quad 2 C-2 D=0,$$
                    $$A=-frac{2}{5},quad B=-frac{4}{5},quad C=-frac{5}{4},quad D=-frac{5}{4},$$
                    $$y_p={{e}^{4 x}}, left( -frac{2 sin{(x)}}{5}-frac{4 cos{(x)}}{5}right) +{{e}^{2 x}}, left( -frac{5 x}{4}-frac{5}{4}right) x\=
                    -frac{2 {{e}^{4 x}}, left( sin{(x)}+2 cos{(x)}right) }{5}-frac{5 {{e}^{2 x}} x, left( x+1right) }{4},$$
                    General solution is
                    $$y=c_1e^{2x}+c_2e^{4x}+y_p$$






                    share|cite|improve this answer









                    $endgroup$



                    Particular solution we find with method of undermined coefficients
                    $$y_p={{e}^{4 x}}, left( A sin{(x)}+B cos{(x)}right) +{{e}^{2 x}} x, left( C x+Dright) $$
                    We get
                    $${{e}^{4 x}}, left( -2 B sin{(x)}-A sin{(x)}-B cos{(x)}+2 A cos{(x)}right) +{{e}^{2 x}}, left( -4 C x-2 D+2 Cright)\ =2 {{e}^{4 x}} sin{(x)}+5 {{e}^{2 x}} x,$$
                    $$-2 B-A=2,quad 2 A-B=0,quad -4 C=5,quad 2 C-2 D=0,$$
                    $$A=-frac{2}{5},quad B=-frac{4}{5},quad C=-frac{5}{4},quad D=-frac{5}{4},$$
                    $$y_p={{e}^{4 x}}, left( -frac{2 sin{(x)}}{5}-frac{4 cos{(x)}}{5}right) +{{e}^{2 x}}, left( -frac{5 x}{4}-frac{5}{4}right) x\=
                    -frac{2 {{e}^{4 x}}, left( sin{(x)}+2 cos{(x)}right) }{5}-frac{5 {{e}^{2 x}} x, left( x+1right) }{4},$$
                    General solution is
                    $$y=c_1e^{2x}+c_2e^{4x}+y_p$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 27 at 7:46









                    Aleksas DomarkasAleksas Domarkas

                    1,54817




                    1,54817












                    • $begingroup$
                      +1 for the detailed calculations. However, you should not "undermine" the "undetermined" coefficients.
                      $endgroup$
                      – farruhota
                      Jan 27 at 10:03


















                    • $begingroup$
                      +1 for the detailed calculations. However, you should not "undermine" the "undetermined" coefficients.
                      $endgroup$
                      – farruhota
                      Jan 27 at 10:03
















                    $begingroup$
                    +1 for the detailed calculations. However, you should not "undermine" the "undetermined" coefficients.
                    $endgroup$
                    – farruhota
                    Jan 27 at 10:03




                    $begingroup$
                    +1 for the detailed calculations. However, you should not "undermine" the "undetermined" coefficients.
                    $endgroup$
                    – farruhota
                    Jan 27 at 10:03


















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