Find general solution for the differential equation $x^3y^{'''}+x^2y^{''}+3xy^{'}-8y=0$ Find general solution for the differential equation $x^3y^{'''}+x^2y^{''}+3xy^{'}-8y=0$
This is the Euler differential equation which can be solved by substitution $x=e^t$. I don't understand the following differential relations:
$$xy^{'}=x\frac{dy}{dx}=\frac{dy}{dt}
$$
$$x^2y^{''}=x^2\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{d}{dt}-1\right)y
$$
$$x^3y^{'''}=x^3\frac{d^3y}{dx^3}=\frac{d}{dt}\left(\frac{d}{dt}-1\right)\left(\frac{d}{dt}-2\right)y
$$
How to evaluate these relations?
From here, it is easy to solve the equation, which is homogeneous with constant coefficients.
General solution is $y=c_1e^{2\ln x}+c_2e^{-i2\ln x}+c_3e^{i2\ln x}$
 A: HINT:
$$x^3y'''(x)+x^2y''(x)+3xy'(x)-8y(x)=0\Longleftrightarrow$$
$$x^3\cdot\frac{\text{d}^3y(x)}{\text(d)x^3}+x^2\cdot\frac{\text{d}^2y(x)}{\text(d)x^2}+3x\cdot\frac{\text{d}y(x)}{\text(d)x}-8y(x)=0\Longleftrightarrow$$

Assume the solution will be proportional to $x^{\lambda}$ for some constant $\lambda$.
Substitute $y(x)=x^{\lambda}$:

$$x^3\cdot\frac{\text{d}^3x^{\lambda}}{\text(d)x^3}+x^2\cdot\frac{\text{d}^2x^{\lambda}}{\text(d)x^2}+3x\cdot\frac{\text{d}x^{\lambda}}{\text(d)x}-8x^{\lambda}=0\Longleftrightarrow$$

Substitute $\frac{\text{d}x^{\lambda}}{\text{d}x}=\lambda x^{\lambda-1}$:

$$\lambda^3x^{\lambda}-2\lambda^2x^{\lambda}+4\lambda x^{\lambda}-8x^{\lambda}=0\Longleftrightarrow$$
$$x^{\lambda}\left(\lambda^3-2\lambda^2+4\lambda-8\right)=0\Longleftrightarrow$$

Assuming $x\ne 0$, the zeros must come from the polynomial:

$$\lambda^3-2\lambda^2+4\lambda-8=0\Longleftrightarrow$$
$$\left(\lambda-2\right)\left(\lambda^2+4\right)=0$$
A: It is called Cauchy's equation; what you did was correct. Set the operator $D: = \frac {d} {dt}$. Then form the auxiliary equation and solve it. The equation will be of the form $$(D^3-2D^2+4D-8)y=0,$$ where $x = e^t$. This gives $D = 2, \pm 2i$.
