Find the general form for solutions of x 2y ′′ − 4xy′ + 6y = 0. (Hint: look for solutions of form y = x r .)

We are tasked with solving the second-order linear differential equation:

\[
x^2 y'' - 4x y' + 6y = 0
\]

The hint suggests looking for solutions of the form \( y = x^r \), where \( r \) is a constant to be determined. Let's proceed step by step:

### Step 1: Compute the derivatives of \( y = x^r \).

First, we compute the first and second derivatives of \( y = x^r \):

- First derivative:
\[
y' = \frac{d}{dx}(x^r) = r x^{r-1}
\]

- Second derivative:
\[
y'' = \frac{d}{dx}(r x^{r-1}) = r(r-1) x^{r-2}
\]

### Step 2: Substitute into the differential equation.

Substitute \( y = x^r \), \( y' = r x^{r-1} \), and \( y'' = r(r-1) x^{r-2} \) into the original equation:

\[
x^2 \cdot r(r-1) x^{r-2} - 4x \cdot r x^{r-1} + 6x^r = 0
\]

Simplifying each term:

- The first term: \( x^2 \cdot r(r-1) x^{r-2} = r(r-1) x^r \)
- The second term: \( -4x \cdot r x^{r-1} = -4r x^r \)
- The third term: \( 6x^r = 6x^r \)

Thus, the equation becomes:

\[
r(r-1) x^r - 4r x^r + 6x^r = 0
\]

### Step 3: Factor out \( x^r \).

Since \( x^r \) appears in all terms, we can factor it out (assuming \( x \neq 0 \)):

\[
x^r \left[ r(r-1) - 4r + 6 \right] = 0
\]

Now, solve the quadratic equation in \( r \):

\[
r(r-1) - 4r + 6 = 0
\]

Expanding:

\[
r^2 - r - 4r + 6 = 0
\]

Simplifying:

\[
r^2 - 5r + 6 = 0
\]

### Step 4: Solve the quadratic equation.

Solve the quadratic equation:

\[
r^2 - 5r + 6 = 0
\]

The solutions are:

\[
r = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)} = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}
\]

Thus, the two roots are:

\[
r_1 = \frac{5 + 1}{2} = 3, \quad r_2 = \frac{5 - 1}{2} = 2
\]

### Step 5: General solution.

Since the equation is linear and homogeneous, the general solution is a linear combination of the two independent solutions \( y = x^r \) corresponding to \( r_1 = 3 \) and \( r_2 = 2 \):

\[
y(x) = C_1 x^3 + C_2 x^2
\]

where \( C_1 \) and \( C_2 \) are arbitrary constants.

### Final Answer:
The general form of the solution to the differential equation is:

\[
y(x) = C_1 x^3 + C_2 x^2
\]

Would you like more details or have any questions?

### Related Questions:
1. How would the solution change if the right-hand side of the equation were not zero (non-homogeneous)?
2. What if the solution form \( y = x^r \) was not suggested? How would we solve the equation?
3. How can we verify that the solution satisfies the original differential equation?
4. What is the significance of the roots \( r_1 \) and \( r_2 \) in terms of the solution's behavior?
5. Can this method be applied to any second-order differential equation of this type?

### Tip:
For second-order linear differential equations with variable coefficients, looking for power solutions (such as \( y = x^r \)) is often a very effective approach.

Find the general form for solutions of x^2 y'' - 4xy' + 6y = 0. (Hint: look for solutions of form y = x^r.)

Discover the step-by-step solution to the second-order linear differential equation x^2 y'' - 4xy' + 6y = 0. This guide covers the method of solving using power solutions (y = x^r) and includes detailed derivations and quadratic equation solving. Ideal for undergraduate students studying differential equations.

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