Answer to Question #211910 in Differential Equations for Natchathiran

Question

x^2d^2y/dx^2-xdy/dx+y=log x+sin(log x)+1/x

Answer 1

$$x^2y''-xy'+y=\ln x+\sin(\ln x)+\dfrac{1}{x}$$

Let $x=e^t.$ Then

$$y'=e^{-t}y'_t, y''=-e^{-2t}y'_t+e^{-2t}y''_{t^2}$$

$$\ln x=t$$

$$e^{2t}(-e^{-2t}y'_t+e^{-2t}y''_{t^2})-e^te^{-t}y'_t+y=t+\sin t+e^{-t}$$

$$y''_{t^2}-2y_t'+y=t+\sin t+e^{-t}$$

Write the related homogeneous or complementary equation:

$$y''_{t^2}-2y_t'+y=0$$

The general solution of a nonhomogeneous equation is the sum of the general solution $y_h(x)$ of the related homogeneous equation and a particular solution $y_p(x)$ of the nonhomogeneous equation:

$$y(x)=y_h(x)+y_p(x)$$

Consider a homogeneous equation 

$$y''_{t^2}-2y_t'+y=0$$

Write the characteristic (auxiliary) equation:

$$r^2-2r+1=0$$

$$(r-1)^2=0$$

$$r_1=1, r_2=1$$

The general solution of the homogeneous equation is

$$y_h(t)=C_1e^t+C_2te^t$$

Let

$$y_p=At+B+C\sin t+D\cos t+Ee^{-t}$$

Then

$$y_p'=A+C\cos t-D\sin t-Ee^{-t}$$

$$y_p''=-C\sin t-D\cos t+Ee^{-t}$$

Substitute

$$-C\sin t-D\cos t+Ee^{-t}$$

$$-2A+2C\cos t+2D\sin t+2Ee^{-t}$$

$$+At+B+C\sin t+D\cos t+Ee^{-t}$$

$$=t+\sin t+e^{-t}$$

$$A=1, B=2$$

$$C=0, D=\dfrac{1}{2}, E=\dfrac{1}{4}$$

The general solution of a second order homogeneous differential equation be

$$y(t)=C_1e^t+C_2te^t+t+2+\dfrac{1}{2}\cos t+\dfrac{1}{4}e^{-t}$$

$$y(x)=C_1x+C_2x\ln x+\ln x+2+\dfrac{1}{2}\cos (\ln x)+\dfrac{1}{4x}$$