Answer to Question #188246 in Differential Equations for Akansha sisodia

Question

(D²+7DD'+12D'²)z=sinhx

Answer 1

$(D^2+7DD'+12D'^2)z=sinhx$

Auxilary equation is-

  $m^2+7m+12=0\\ \Rightarrow(m+3)(m+4)=0\\ \Rightarrow m=-3,-4$

Complimentary function CF is-

       $CF= f_1(y-3x)+f_2(y-4x)$

Then Paticular Integral-

         $PI=\dfrac{1}{D^2+7DD'+12D'^2}sinhx$

     

           $=\dfrac{1}{D^2+7DD'+12D'^2}(\dfrac{e^x-e^{-x}}{2})$

           $=\dfrac{1}{2}(\dfrac{e^x}{D^2+7DD'+12D'^2}-\dfrac{e^{-x}}{D^2+7DD'+12D'^2})$

          $=\dfrac{1}{2}(\dfrac{e^x}{(1^2+7(1)+12(0)}-\dfrac{e^{-x}}{(-1)^2+7(1)+12(0)})$

         $=\dfrac{1}{2}(\dfrac{e^x-e^{-x}}{8})$

         $=\dfrac{sinhx}{8}$

  Solution is-

 $z=f_1(y+3x)+f_2(y+6x)+\dfrac{sinhx}{8}$