Answer to Question #222706 in Differential Equations for sam

Question

x²y''+xy'+4y=2xlnx

Answer 1

$$x^2y''+xy'+4y=2x\ln x$$

Homogeneous equation

$$x^2y''+xy'+4y=0$$

Let $y=x^r$

$y'=rx^{r-1}$

$y''=r(r-1)x^{r-2}$

$$r(r-1)+r+4=0$$

$$r=\pm2 i$$

The general solution of the homogeneous equation is

$$y_h=C_1\cos(2\ln x)+C_2\sin(2\ln x)$$

$$x^2y''+xy'+4y=2x\ln x$$

Divide both sides by $x^2$

$$y''+\dfrac{1}{x}y'+\dfrac{4}{x^2}y=\dfrac{2}{x}\ln x=>g(x)=\dfrac{2}{x}\ln x$$

Wronskian

$$W(y_1, y_2)=\begin{vmatrix} \cos(2\ln x)& \sin(2\ln x) \\ - \dfrac{2}{x}\sin (2\ln x) & \dfrac{2}{x}\cos(2\ln x) \end{vmatrix}=\dfrac{2}{x}$$

$$W_1=\begin{vmatrix} 0& \sin(2\ln x) \\ \dfrac{2}{x}\ln x & \dfrac{2}{x}\cos(2\ln x) \end{vmatrix}=-\dfrac{2}{x}\ln x \sin(2\ln x)$$

$$W_2=\begin{vmatrix} \cos(2\ln x) & 0 \\ - \dfrac{2}{x}\sin (2\ln x) & \dfrac{2}{x}\ln x \end{vmatrix}=\dfrac{2}{x}\ln x \cos(2\ln x)$$

$$u_1=\int\dfrac{W_1}{W}dx=\int\dfrac{-\dfrac{2}{x}\ln x \sin(2\ln x)}{\dfrac{2}{x}}dx$$

$$=-\int \ln x \sin(2\ln x)dx$$

$$=-\dfrac{1}{25}x(5\ln x+3)\sin(2\ln x)+\dfrac{1}{25}(10\ln x-4)\cos(2\ln x)$$

$$u_2=\int\dfrac{W_2}{W}dx=\int\dfrac{\dfrac{2}{x}\ln x \cos(2\ln x)}{\dfrac{2}{x}}dx$$

$$=\int \ln x \cos(2\ln x)dx$$

$$=\dfrac{1}{25}x(10\ln x-4)\sin(2\ln x)+\dfrac{1}{25}(5\ln x+3)\cos(2\ln x)$$

Particular solution of the non-homogeneous equation is

$$y_p=u_1y_1+u_2y_2$$

$$y_p=-\dfrac{1}{25}x(5\ln x+3)\sin(2\ln x)\cos(2\ln x)$$

$$+\dfrac{1}{25}(10\ln x-4)\cos^2(2\ln x)$$

$$+\dfrac{1}{25}x(10\ln x-4)\sin^2(2\ln x)$$

$$+\dfrac{1}{25}(5\ln x+3)\sin(2\ln x)\cos(2\ln x)$$

The general solution of the homogeneous equation is

$$y=C_1\cos(2\ln x)+C_2\sin(2\ln x)$$

$$-\dfrac{1}{25}x(5\ln x+3)\sin(2\ln x)\cos(2\ln x)$$

$$+\dfrac{1}{25}(10\ln x-4)\cos^2(2\ln x)$$

$$+\dfrac{1}{25}x(10\ln x-4)\sin^2(2\ln x)$$

$$+\dfrac{1}{25}(5\ln x+3)\sin(2\ln x)\cos(2\ln x)$$