Question

Help me with this
x^2*d^3y/dx^3+2*dy/dx-2/x*y=x*ln(x)+3

Answer 1

can you write it a little clearer; maybe use the draw?

Answer 2

You have the Euler ODE
\[x^2\frac{d^3y}{dx^3}+2\frac{dy}{dx}-\frac{2}{x}y=x\ln x+3\]or perhaps it doesn't look like the usual form, but we can easily obtain it by multiplying throughout by \(x\):
\[x^3\frac{d^3y}{dx^3}+2x\frac{dy}{dx}-2y=x^2\ln x+3x\]
To begin with, consider the substitution \(x=e^t\). Call \(z\) our new dependent variable a function of a new independent variable \(t\):
\[y(x)=y(e^t)=z(t)=z(\ln x)\]with derivatives
\[\begin{align*}\frac{dy}{dx}&=\frac{1}{x}\frac{dz}{dt}\\[1ex]
\frac{d^2y}{dx^2}&=\frac{1}{x^2}\left(\frac{d^2z}{dt^2}-\frac{dz}{dt}\right)\\[1ex]
\frac{d^3y}{dx^3}&=\frac{1}{x^3}\left(\frac{d^3z}{dt^3}-3\frac{d^2z}{dt^2}+2\frac{dz}{dt}\right)\end{align*}\]
Substitute these into the ODE:
\[\begin{align*}\left(\frac{d^3z}{dt^3}-3\frac{d^2z}{dt^2}+2\frac{dz}{dt}\right)+2\frac{dz}{dt}-2z&=te^{2t}+3e^t\\[1ex]
\frac{d^3z}{dt^3}-3\frac{d^2z}{dt^2}+4\frac{dz}{dt}-2z&=te^{2t}+3e^t\\[1ex]
\end{align*}\]
That should make things easier, I hope?