Differential equation - delta function












-1












$begingroup$


How to solve this equation please?



$-y''+2ay'+(b^2-a^2)y=delta$



I found charakteristic roots $lambda_{1,2} = a pm sqrt{a^2+(b^2-a^2)}$.



Thank you






ordinary-differential-equations







share|cite|improve this question









$endgroup$

















    -1












    $begingroup$


    How to solve this equation please?



    $-y''+2ay'+(b^2-a^2)y=delta$



    I found charakteristic roots $lambda_{1,2} = a pm sqrt{a^2+(b^2-a^2)}$.



    Thank you






    ordinary-differential-equations







    share|cite|improve this question









    $endgroup$















      -1












      -1








      -1


      0



      $begingroup$


      How to solve this equation please?



      $-y''+2ay'+(b^2-a^2)y=delta$



      I found charakteristic roots $lambda_{1,2} = a pm sqrt{a^2+(b^2-a^2)}$.



      Thank you






      ordinary-differential-equations







      share|cite|improve this question









      $endgroup$




      How to solve this equation please?



      $-y''+2ay'+(b^2-a^2)y=delta$



      I found charakteristic roots $lambda_{1,2} = a pm sqrt{a^2+(b^2-a^2)}$.



      Thank you






      ordinary-differential-equations




      ordinary-differential-equations






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 6 at 9:35









      ElisabethElisabeth

      1066




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          1 Answer
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          $begingroup$

          The solution has the form $y=y_0u$ where $u$ is the step function, $y_0$ a homogeneous solution with $y_0(0)=0$ and $y_0'(0)=-1$.



          Also, you should simplify your expressions, $λ_{1,2}=a±b$.



          Thus the solution for $y_0$ is $y_0(t)=Ce^{ax}sinh(bx)$ with
          $$y_0'(x)=Ce^{ax}(asinh(bx)+bcosh(bx))implies Cb=-1, ~~C=-frac1b.$$
          So that
          $$
          y(x)=-frac{e^{ax}sinh(bx)}b u(x)
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you, so I have $y^+ = c_1^+e^{-(a+b)x}+c_2^+e^{-(a-b)x}$ for x > 0 and $y^- = c_1^-e^{-(a+b)x}+c_2^-e^{-(a-b)x}$ . What will be please the conditions of sticking?
            $endgroup$
            – Elisabeth
            Jan 6 at 9:57






          • 1




            $begingroup$
            Usually, if no further information is given, the implicit condition for this type of problem is that $y(t)=0$ for $t<0$, conforming to the use of the Laplace transform for the solution. Thus $c_{1,2}^-=0$.
            $endgroup$
            – LutzL
            Jan 6 at 10:05












          • $begingroup$
            @Elisabeth : Check again the sign in the exponential factors.
            $endgroup$
            – LutzL
            Jan 6 at 10:09










          • $begingroup$
            There should be + in the arguments of first exponentials functions. How to do Laplace transform in this equation? I can't find any solved example
            $endgroup$
            – Elisabeth
            Jan 6 at 10:31






          • 1




            $begingroup$
            You look up a Laplace table like tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx and apply term-by-term, $Y={cal L}[y]$ has the equation $$-(s^2Y(s)-sy(0)-y'(0))+2a(sY(s)-y(0))+(b^2-a^2)Y(s)=1,$$ then isolate $$Y(s)=-frac{1+2ay(0)-y'(0)-sy(0)}{(s-a)^2-b^2}$$ and then perform a partial fraction decomposition or if possible look up the terms for the inverse Laplace transform by reading the table backwards.
            $endgroup$
            – LutzL
            Jan 6 at 11:04













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          $begingroup$

          The solution has the form $y=y_0u$ where $u$ is the step function, $y_0$ a homogeneous solution with $y_0(0)=0$ and $y_0'(0)=-1$.



          Also, you should simplify your expressions, $λ_{1,2}=a±b$.



          Thus the solution for $y_0$ is $y_0(t)=Ce^{ax}sinh(bx)$ with
          $$y_0'(x)=Ce^{ax}(asinh(bx)+bcosh(bx))implies Cb=-1, ~~C=-frac1b.$$
          So that
          $$
          y(x)=-frac{e^{ax}sinh(bx)}b u(x)
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you, so I have $y^+ = c_1^+e^{-(a+b)x}+c_2^+e^{-(a-b)x}$ for x > 0 and $y^- = c_1^-e^{-(a+b)x}+c_2^-e^{-(a-b)x}$ . What will be please the conditions of sticking?
            $endgroup$
            – Elisabeth
            Jan 6 at 9:57






          • 1




            $begingroup$
            Usually, if no further information is given, the implicit condition for this type of problem is that $y(t)=0$ for $t<0$, conforming to the use of the Laplace transform for the solution. Thus $c_{1,2}^-=0$.
            $endgroup$
            – LutzL
            Jan 6 at 10:05












          • $begingroup$
            @Elisabeth : Check again the sign in the exponential factors.
            $endgroup$
            – LutzL
            Jan 6 at 10:09










          • $begingroup$
            There should be + in the arguments of first exponentials functions. How to do Laplace transform in this equation? I can't find any solved example
            $endgroup$
            – Elisabeth
            Jan 6 at 10:31






          • 1




            $begingroup$
            You look up a Laplace table like tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx and apply term-by-term, $Y={cal L}[y]$ has the equation $$-(s^2Y(s)-sy(0)-y'(0))+2a(sY(s)-y(0))+(b^2-a^2)Y(s)=1,$$ then isolate $$Y(s)=-frac{1+2ay(0)-y'(0)-sy(0)}{(s-a)^2-b^2}$$ and then perform a partial fraction decomposition or if possible look up the terms for the inverse Laplace transform by reading the table backwards.
            $endgroup$
            – LutzL
            Jan 6 at 11:04


















          1












          $begingroup$

          The solution has the form $y=y_0u$ where $u$ is the step function, $y_0$ a homogeneous solution with $y_0(0)=0$ and $y_0'(0)=-1$.



          Also, you should simplify your expressions, $λ_{1,2}=a±b$.



          Thus the solution for $y_0$ is $y_0(t)=Ce^{ax}sinh(bx)$ with
          $$y_0'(x)=Ce^{ax}(asinh(bx)+bcosh(bx))implies Cb=-1, ~~C=-frac1b.$$
          So that
          $$
          y(x)=-frac{e^{ax}sinh(bx)}b u(x)
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you, so I have $y^+ = c_1^+e^{-(a+b)x}+c_2^+e^{-(a-b)x}$ for x > 0 and $y^- = c_1^-e^{-(a+b)x}+c_2^-e^{-(a-b)x}$ . What will be please the conditions of sticking?
            $endgroup$
            – Elisabeth
            Jan 6 at 9:57






          • 1




            $begingroup$
            Usually, if no further information is given, the implicit condition for this type of problem is that $y(t)=0$ for $t<0$, conforming to the use of the Laplace transform for the solution. Thus $c_{1,2}^-=0$.
            $endgroup$
            – LutzL
            Jan 6 at 10:05












          • $begingroup$
            @Elisabeth : Check again the sign in the exponential factors.
            $endgroup$
            – LutzL
            Jan 6 at 10:09










          • $begingroup$
            There should be + in the arguments of first exponentials functions. How to do Laplace transform in this equation? I can't find any solved example
            $endgroup$
            – Elisabeth
            Jan 6 at 10:31






          • 1




            $begingroup$
            You look up a Laplace table like tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx and apply term-by-term, $Y={cal L}[y]$ has the equation $$-(s^2Y(s)-sy(0)-y'(0))+2a(sY(s)-y(0))+(b^2-a^2)Y(s)=1,$$ then isolate $$Y(s)=-frac{1+2ay(0)-y'(0)-sy(0)}{(s-a)^2-b^2}$$ and then perform a partial fraction decomposition or if possible look up the terms for the inverse Laplace transform by reading the table backwards.
            $endgroup$
            – LutzL
            Jan 6 at 11:04
















          1












          1








          1





          $begingroup$

          The solution has the form $y=y_0u$ where $u$ is the step function, $y_0$ a homogeneous solution with $y_0(0)=0$ and $y_0'(0)=-1$.



          Also, you should simplify your expressions, $λ_{1,2}=a±b$.



          Thus the solution for $y_0$ is $y_0(t)=Ce^{ax}sinh(bx)$ with
          $$y_0'(x)=Ce^{ax}(asinh(bx)+bcosh(bx))implies Cb=-1, ~~C=-frac1b.$$
          So that
          $$
          y(x)=-frac{e^{ax}sinh(bx)}b u(x)
          $$






          share|cite|improve this answer











          $endgroup$



          The solution has the form $y=y_0u$ where $u$ is the step function, $y_0$ a homogeneous solution with $y_0(0)=0$ and $y_0'(0)=-1$.



          Also, you should simplify your expressions, $λ_{1,2}=a±b$.



          Thus the solution for $y_0$ is $y_0(t)=Ce^{ax}sinh(bx)$ with
          $$y_0'(x)=Ce^{ax}(asinh(bx)+bcosh(bx))implies Cb=-1, ~~C=-frac1b.$$
          So that
          $$
          y(x)=-frac{e^{ax}sinh(bx)}b u(x)
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 11:12

























          answered Jan 6 at 9:43









          LutzLLutzL

          59.3k42057




          59.3k42057












          • $begingroup$
            Thank you, so I have $y^+ = c_1^+e^{-(a+b)x}+c_2^+e^{-(a-b)x}$ for x > 0 and $y^- = c_1^-e^{-(a+b)x}+c_2^-e^{-(a-b)x}$ . What will be please the conditions of sticking?
            $endgroup$
            – Elisabeth
            Jan 6 at 9:57






          • 1




            $begingroup$
            Usually, if no further information is given, the implicit condition for this type of problem is that $y(t)=0$ for $t<0$, conforming to the use of the Laplace transform for the solution. Thus $c_{1,2}^-=0$.
            $endgroup$
            – LutzL
            Jan 6 at 10:05












          • $begingroup$
            @Elisabeth : Check again the sign in the exponential factors.
            $endgroup$
            – LutzL
            Jan 6 at 10:09










          • $begingroup$
            There should be + in the arguments of first exponentials functions. How to do Laplace transform in this equation? I can't find any solved example
            $endgroup$
            – Elisabeth
            Jan 6 at 10:31






          • 1




            $begingroup$
            You look up a Laplace table like tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx and apply term-by-term, $Y={cal L}[y]$ has the equation $$-(s^2Y(s)-sy(0)-y'(0))+2a(sY(s)-y(0))+(b^2-a^2)Y(s)=1,$$ then isolate $$Y(s)=-frac{1+2ay(0)-y'(0)-sy(0)}{(s-a)^2-b^2}$$ and then perform a partial fraction decomposition or if possible look up the terms for the inverse Laplace transform by reading the table backwards.
            $endgroup$
            – LutzL
            Jan 6 at 11:04




















          • $begingroup$
            Thank you, so I have $y^+ = c_1^+e^{-(a+b)x}+c_2^+e^{-(a-b)x}$ for x > 0 and $y^- = c_1^-e^{-(a+b)x}+c_2^-e^{-(a-b)x}$ . What will be please the conditions of sticking?
            $endgroup$
            – Elisabeth
            Jan 6 at 9:57






          • 1




            $begingroup$
            Usually, if no further information is given, the implicit condition for this type of problem is that $y(t)=0$ for $t<0$, conforming to the use of the Laplace transform for the solution. Thus $c_{1,2}^-=0$.
            $endgroup$
            – LutzL
            Jan 6 at 10:05












          • $begingroup$
            @Elisabeth : Check again the sign in the exponential factors.
            $endgroup$
            – LutzL
            Jan 6 at 10:09










          • $begingroup$
            There should be + in the arguments of first exponentials functions. How to do Laplace transform in this equation? I can't find any solved example
            $endgroup$
            – Elisabeth
            Jan 6 at 10:31






          • 1




            $begingroup$
            You look up a Laplace table like tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx and apply term-by-term, $Y={cal L}[y]$ has the equation $$-(s^2Y(s)-sy(0)-y'(0))+2a(sY(s)-y(0))+(b^2-a^2)Y(s)=1,$$ then isolate $$Y(s)=-frac{1+2ay(0)-y'(0)-sy(0)}{(s-a)^2-b^2}$$ and then perform a partial fraction decomposition or if possible look up the terms for the inverse Laplace transform by reading the table backwards.
            $endgroup$
            – LutzL
            Jan 6 at 11:04


















          $begingroup$
          Thank you, so I have $y^+ = c_1^+e^{-(a+b)x}+c_2^+e^{-(a-b)x}$ for x > 0 and $y^- = c_1^-e^{-(a+b)x}+c_2^-e^{-(a-b)x}$ . What will be please the conditions of sticking?
          $endgroup$
          – Elisabeth
          Jan 6 at 9:57




          $begingroup$
          Thank you, so I have $y^+ = c_1^+e^{-(a+b)x}+c_2^+e^{-(a-b)x}$ for x > 0 and $y^- = c_1^-e^{-(a+b)x}+c_2^-e^{-(a-b)x}$ . What will be please the conditions of sticking?
          $endgroup$
          – Elisabeth
          Jan 6 at 9:57




          1




          1




          $begingroup$
          Usually, if no further information is given, the implicit condition for this type of problem is that $y(t)=0$ for $t<0$, conforming to the use of the Laplace transform for the solution. Thus $c_{1,2}^-=0$.
          $endgroup$
          – LutzL
          Jan 6 at 10:05






          $begingroup$
          Usually, if no further information is given, the implicit condition for this type of problem is that $y(t)=0$ for $t<0$, conforming to the use of the Laplace transform for the solution. Thus $c_{1,2}^-=0$.
          $endgroup$
          – LutzL
          Jan 6 at 10:05














          $begingroup$
          @Elisabeth : Check again the sign in the exponential factors.
          $endgroup$
          – LutzL
          Jan 6 at 10:09




          $begingroup$
          @Elisabeth : Check again the sign in the exponential factors.
          $endgroup$
          – LutzL
          Jan 6 at 10:09












          $begingroup$
          There should be + in the arguments of first exponentials functions. How to do Laplace transform in this equation? I can't find any solved example
          $endgroup$
          – Elisabeth
          Jan 6 at 10:31




          $begingroup$
          There should be + in the arguments of first exponentials functions. How to do Laplace transform in this equation? I can't find any solved example
          $endgroup$
          – Elisabeth
          Jan 6 at 10:31




          1




          1




          $begingroup$
          You look up a Laplace table like tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx and apply term-by-term, $Y={cal L}[y]$ has the equation $$-(s^2Y(s)-sy(0)-y'(0))+2a(sY(s)-y(0))+(b^2-a^2)Y(s)=1,$$ then isolate $$Y(s)=-frac{1+2ay(0)-y'(0)-sy(0)}{(s-a)^2-b^2}$$ and then perform a partial fraction decomposition or if possible look up the terms for the inverse Laplace transform by reading the table backwards.
          $endgroup$
          – LutzL
          Jan 6 at 11:04






          $begingroup$
          You look up a Laplace table like tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx and apply term-by-term, $Y={cal L}[y]$ has the equation $$-(s^2Y(s)-sy(0)-y'(0))+2a(sY(s)-y(0))+(b^2-a^2)Y(s)=1,$$ then isolate $$Y(s)=-frac{1+2ay(0)-y'(0)-sy(0)}{(s-a)^2-b^2}$$ and then perform a partial fraction decomposition or if possible look up the terms for the inverse Laplace transform by reading the table backwards.
          $endgroup$
          – LutzL
          Jan 6 at 11:04




















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