Answer to Question #195974 in Differential Equations for Khan

Question

Y"+6y'+9y=-xe^4x by variation of parameters

Answer 1

$$y''+6y'+9y=-xe^{4x}$$

The corresponding homogeneous equation is

$$y''+6y'+9y=0$$

The auxiliary (characteristic) equation is given by 

$$r^2+6r+9=0$$

$$(r+3)^2=0$$

There is one repeated real root $r=-3.$

The general solution of the homogeneous equation is

$$y_c=C_1e^{-3x}+C_2xe^{-3x}$$

We have

$$y_1=e^{-3x}, y_2=xe^{-3x}, g(x)=-xe^{4x}$$

The Wronskian of these two functions is

$$W(y_1,y_2)=\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}=\begin{vmatrix} e^{-3x} & xe^{-3x} \\ -3e^{-3x} & (1-3x)e^{-3x} \end{vmatrix}$$

$$=e^{-3x}(1-3x)e^{-3x}-xe^{-3x}(-3)e^{-3x}$$

$$=e^{-6x}(1-3x+3x)=e^{-6x}$$

$$W_1=\begin{vmatrix} 0 & y_2 \\ g(x) & y_2' \end{vmatrix}=\begin{vmatrix} 0 & xe^{-3x} \\ -xe^{4x} & (1-3x)e^{-3x} \end{vmatrix}=x^2e^x$$

$$W_2=\begin{vmatrix} y_1 & 0 \\ y_1' & g(x) \end{vmatrix}=\begin{vmatrix} e^{-3x} & 0 \\ -3e^{-3x} & -xe^{4x} \end{vmatrix}=xe^x$$

$$u_1=\int\dfrac{W_1}{W}dx=\int\dfrac{x^2e^x}{e^{-6x}}dx=\int x^2e^{7x}dx$$

$$=\dfrac{1}{7}x^2e^{7x}-\dfrac{2}{7}\int xe^{7x}dx$$

$$=\dfrac{1}{7}x^2e^{7x}-\dfrac{2}{49}xe^{7x}+\dfrac{2}{49}\int e^{7x}dx$$

$$=\dfrac{1}{7}x^2e^{7x}-\dfrac{2}{49}xe^{7x}+\dfrac{2}{343}e^{7x}$$

$$u_2=\int\dfrac{W_2}{W}dx=\int\dfrac{xe^x}{e^{-6x}}dx=\int xe^{7x}dx$$

$$=\dfrac{1}{7}xe^{7x}-\dfrac{1}{7}\int e^{7x}dx$$

$$=\dfrac{1}{7}xe^{7x}-\dfrac{1}{49}e^{7x}$$

$$y_p=u_1y_1+u_2y_2$$

$$=(\dfrac{1}{7}x^2e^{7x}-\dfrac{2}{49}xe^{7x}+\dfrac{2}{343}e^{7x})(e^{-3x})+$$

$$+(\dfrac{1}{7}xe^{7x}-\dfrac{1}{49}e^{7x})(xe^{-3x})$$

$$=(\dfrac{2}{7}x^2-\dfrac{3}{49}x+\dfrac{2}{343})e^{4x}$$

The general solution of the homogeneous equation is

$$y=y_c+y_p$$

$$=C_1e^{-3x}+C_2xe^{-3x}+\big(\dfrac{2}{7}x^2-\dfrac{3}{49}x+\dfrac{2}{343}\big)e^{4x}$$