I have
$$
x^2\frac{d^{2}y}{dx^2} – 3x\frac{dy}{dx} +3y = x^2 – 3x + 3 + 3\ln{x}
$$
the general procedure
$$
ax^2\frac{d^{2}y}{dx^2} + bx\frac{dy}{dx} + cy = g(x)
$$
taking $x = e^t$
I got the complementary function
$$
Y_c= Ae^t + B*e^{3t}
$$
I got stuck here.
The nonhomogeneous equation is
$$\color{red}{
y''- 4y' + 3y = e^{2t} – 3e^t\\
y''- 4y' + 3y = 3 + 3t}
$$
What I need to do next to complete the general solution?
Best Answer
Set $$ y(x)=z(\log x), $$ then $$ y'(x)=z'(\log x)/x, \,\,\, y'(x)=z''(\log x)/x^2-z'(\log x)/x^2. $$ Hence $$ x^2\frac{d^{2}y}{dx^2} - 3x\frac{dy}{dx} +3y = x^2 - 3x + 3 + 3\log{x}, $$ becomes $$ \big(z''(\log x)-z'(\log x)\big)-3z'(\log x)+3z(\log x)=x^2 - 3x + 3 + 3\log{x}, $$ or $$ z''(x)-4z'(x)+3z(x)=\mathrm{e}^{2x} - 3\mathrm{e}^{x} + 3 + 3x. $$ General solution of the homogeneous $z''-4z'+3z=0$ is $$z_h=c_1\mathrm{e}^{x}+c_2\mathrm{e}^{3x}.$$ Special solution of the inhomogeneous is of the form $$ z_i(x)=a+bx+cx\mathrm{e}^x+d\mathrm{e}^{2x}, $$ for suitable $a,b,c,d$.
Hence $$ y(x)=z(\log x)=\big(a+b\log x+cx\log x+dx^2\big)+\big(c_1x+c_2x^3\big). $$