Answer to Question #222717 in Differential Equations for wesley

Question

y''-2y'-8y=4e^2x-21e^-3x

Answer 1

Solution;

Auxiliary equation is ;

$m^2-2m-8=0$

$m^2-4m+2m-8=0$

$m(m-4)+2(m-4)=0$

$(m+2)(m-4)=0$

m=-2,4

$y_h=C_1e^{4x}+C_2e^{-2x}$

The particular solution;

$y_1=e^{4x}$

$y_2=e^{-2x}$

$g(x)=4e^{2x}-21e^{-3x}$

$W(y_1,y_2)=det\begin{vmatrix} e^{4x}& e^{-2x} \\ 4e^{4x} & -2e^{-2x} \end{vmatrix}$ $=-2e^{2x}-4e^{2x}=-6e^{2x}$

$W_1=det\begin{vmatrix} 0 & e^{-2x}\\ 4e^{2x}-21e^{-3x} & -2e^{-2x} \end{vmatrix}=-e^{-2x}(4e^{2x}-21e^{-3x})=-1+21e^{-5x}$

$W_2=det\begin{vmatrix} e^{4x} & 0 \\ 4e^{4x} & 4e^{2x}-21e^{-3x} \end{vmatrix}=e^{4x}(4e^{2x}-21e^{-3x})=4e^{6x}-21e^{x}$

Let;

$u_1=\int\frac{W_1}{W}dx=\int\frac{-1+21e^{-5x}}{-6e^{2x}}dx=\frac16\int e^{-2x}dx-\frac{21}6\int e^{-7x}dx$

$u_1=\frac{-e^{-2x}}{12}+\frac{e^{-7x}}{2}$

$u_2=\int\frac{W_2}{W}dx=\int\frac{4e^{6x}-21e^x}{-6e^{2x}}=\frac{-4}6\int e^{4x}dx+\frac72\int e^{-x}dx$

$u_2=-\frac16e^{4x}-\frac72e^{-x}$

Let;

$y_p=u_1y_1+u_2y_2$

$y_p=(\frac{e^{-7x}}{2}-\frac{e^{-2x}}{12})e^{4x}+(-\frac16e^{4x}-\frac72e^{-x})e^{-2x}$

$y_p=-\frac{e^{2x}}{4}-3e^{-3x}$

The general solution is;

$y=y_h+y_p$

$y=C_1e^{4x}+C_2e^{-2x}-\frac{e^{2x}}{4}-3e^{-3x}$