Answer to Question #287618 in Differential Equations for Ali

Question

x^3y''' + 2x^2y'' = x + sin(lnx)

Answer 1

$y''=v,y'''=v'$

$v'+2v/x=(x+sin(lnx))/x^3$

$\mu(x)=e^{\int 2/x dx}=x^2$

multiply both sides by $\mu(x)$ :

$x^2v'+2xv=(x+sin(lnx))/x$

$\frac{d}{dx}(x^2v)=(x+sin(lnx))/x$

$x^2v=\int(x+sin(lnx))/xdx$

$x^2v=x-cos(lnx)+c_1$

$v=y''=(x-cos(lnx)+c_1)/x^2$

$y'=\int(x-cos(lnx)+c_1)/x^2dx$

$\int \frac{cos(lnx)}{x^2}dx=|u=lnx,dx=xdu|=\int e^{-u}cosudu=$

$=-e^{-u}cosu-\intop e^{-u}cosudu=-e^{-u}cosu+-e^{-u}sinu-\intop e^{-u}cosudu$

$\intop e^{-u}cosudu=\frac{cos(lnu)-sin(lnxu)}{2}=\frac{cos(lnx)-sin(lnx)}{2x}$

$y'=\frac{cos(lnx)-sin(lnx)}{2x}+lnx-c_1/x+c_2$

$y=\int (\frac{cos(lnx)-sin(lnx)}{2x}+lnx-c_1/x+c_2)dx$

$\int \frac{cos(lnx)-sin(lnx)}{x}dx=\int (cos(lnx)-sin(lnx))d(lnx)=sin(lnx)+cos(lnx)$

$y=\frac{cos(lnx)+sin(lnx)}{2}-x+xlnx-c_1lnx+c_2x+c_3$