We are tasked with solving the initial value problem (IVP):

$x^2 y'' + 3x y' + y = 0, \quad y(1) = 4, \quad y'(1) = -1.$

Step 1: Verify type of equation

This is a second-order linear differential equation with variable coefficients. It is in standard form for an Euler-Cauchy equation, which takes the form: $x^2 y'' + a x y' + b y = 0.$

Here: $a = 3, \quad b = 1.$

Step 2: Substituting $y = x^r$

We use the substitution $y = x^r$, where $r$ is to be determined. Calculating derivatives: $y' = r x^{r-1}, \quad y'' = r(r-1) x^{r-2}.$

Substitute $y = x^r$ into the equation: [ x^2 \big[r(r-1)x^{r-2}\big] + 3x\big[rx^{r-1}\big] + x^r = 0. ]

Simplify each term: $r(r-1)x^r + 3r x^r + x^r = 0.$

Factor out $x^r$: $x^r \big[r(r-1) + 3r + 1\big] = 0.$

Since $x^r \neq 0$, the characteristic equation is: $r(r-1) + 3r + 1 = 0.$

Step 3: Solve the characteristic equation

Simplify the equation: $r^2 - r + 3r + 1 = 0,$ $r^2 + 2r + 1 = 0.$

Factorize: $(r+1)^2 = 0.$

Thus, $r = -1$ is a repeated root.

Step 4: General solution for repeated roots

For repeated roots, the general solution is: $y(x) = C_1 x^{-1} + C_2 x^{-1} \ln x,$ where $C_1$ and $C_2$ are constants to be determined.

Step 5: Apply initial conditions

The general solution is: $y(x) = \frac{C_1}{x} + \frac{C_2 \ln x}{x}.$

First derivative:

$y'(x) = -\frac{C_1}{x^2} - \frac{C_2 \ln x}{x^2} + \frac{C_2}{x^2}.$

At $x = 1$, the initial conditions are $y(1) = 4$ and $y'(1) = -1$.

1. Apply $y(1) = 4$: $y(1) = \frac{C_1}{1} + \frac{C_2 \ln 1}{1}.$ Since $\ln 1 = 0$, this simplifies to: $C_1 = 4.$

2. Apply $y'(1) = -1$: $y'(1) = -\frac{C_1}{1^2} - \frac{C_2 \ln 1}{1^2} + \frac{C_2}{1^2}.$ Substitute $C_1 = 4$ and $\ln 1 = 0$: $y'(1) = -4 + C_2 = -1.$

Solve for $C_2$: $C_2 = 3.$

Final Solution:

The solution to the IVP is: $y(x) = \frac{4}{x} + \frac{3 \ln x}{x}.$

Let me know if you have questions or want further explanations!

---

Here are 5 related questions to explore:

1. How do we classify Euler-Cauchy equations, and why are they important in differential equations?
2. Can we use a numerical method to solve the given equation and compare the result?
3. How would the solution differ if the characteristic equation had distinct roots?
4. What happens if we replace $\ln x$ with a different function when the roots are repeated?
5. Can you derive the solution if additional external forces (non-homogeneous terms) are added?

Tip: Always check if the differential equation is homogeneous or non-homogeneous before solving!