We are given the second-order linear homogeneous differential equation:

$x^2 y'' - 7x y' + 16y = 0$ with a known solution $y_1(x) = x^4$.

To find a second linearly independent solution $y_2(x)$, we can use the method of reduction of order. This method assumes the second solution has the form:

$y_2(x) = v(x) y_1(x)$

where $v(x)$ is an unknown function to be determined.

Step 1: Substitute $y_2(x) = v(x) y_1(x)$ into the differential equation.

We know that $y_1(x) = x^4$, so we assume $y_2(x) = v(x) x^4$.

Now, we need to compute $y_2'(x)$ and $y_2''(x)$:

• First derivative: $y_2'(x) = v'(x) x^4 + v(x) \cdot 4x^3$

• Second derivative: $y_2''(x) = v''(x) x^4 + 2v'(x) \cdot 4x^3 + v(x) \cdot 12x^2$

Step 2: Substitute $y_2(x), y_2'(x), y_2''(x)$ into the original differential equation.

The original equation is:

$x^2 y'' - 7x y' + 16y = 0$

Substitute the expressions for $y_2''(x), y_2'(x),$ and $y_2(x)$:

$x^2 \left(v''(x) x^4 + 8v'(x) x^3 + 12v(x) x^2 \right) - 7x \left(v'(x) x^4 + 4v(x) x^3 \right) + 16 \cdot v(x) x^4 = 0$

Step 3: Simplify the equation.

Simplify the powers of $x$ in each term:

$x^6 v''(x) + 8x^5 v'(x) + 12x^4 v(x) - 7x^5 v'(x) - 28x^4 v(x) + 16x^4 v(x) = 0$

Combine like terms:

$x^6 v''(x) + (8x^5 - 7x^5) v'(x) + (12x^4 - 28x^4 + 16x^4) v(x) = 0$

This simplifies to:

$x^6 v''(x) + x^5 v'(x) = 0$

Step 4: Solve the resulting equation for $v(x)$.

We can factor out $x^5$ from the equation:

$x^5 \left(x v''(x) + v'(x)\right) = 0$

Since $x^5 \neq 0$ for $x \neq 0$, we are left with:

$x v''(x) + v'(x) = 0$

This is a first-order differential equation in $v'(x)$. To solve this, we can rewrite it as:

$\frac{d}{dx}(v'(x)) + \frac{1}{x} v'(x) = 0$

Step 5: Solve for $v'(x)$.

This equation can be solved by separation of variables. Rewrite it as:

$\frac{dv'(x)}{v'(x)} = -\frac{dx}{x}$

Integrate both sides:

$\ln|v'(x)| = -\ln|x| + C$

Exponentiate both sides:

$v'(x) = \frac{C_1}{x}$

Step 6: Solve for $v(x)$.

Integrate again to find $v(x)$:

$v(x) = C_1 \ln|x| + C_2$

Step 7: Find $y_2(x)$.

Now that we have $v(x)$, we can substitute it back into the expression for $y_2(x)$:

$y_2(x) = v(x) y_1(x) = (C_1 \ln|x| + C_2) x^4$

Since we are looking for a second solution independent from $y_1(x) = x^4$, we can discard the constant term $C_2 x^4$ (as it would simply multiply the known solution). Thus, the second linearly independent solution is:

$y_2(x) = C_1 x^4 \ln|x|$

Therefore, the second solution is:

$y_2(x) = x^4 \ln|x|$

Final Answer:

The second solution to the differential equation is $y_2(x) = x^4 \ln|x|$.

---

Would you like further clarification on any of these steps?

Here are 5 related questions to explore further:

1. How do you determine linear independence between two solutions of a differential equation?
2. What is the general solution to a second-order linear homogeneous differential equation?
3. How does the method of reduction of order work for non-constant coefficients?
4. What is the importance of the Wronskian in determining independent solutions?
5. Can you apply the reduction of order method to a non-homogeneous differential equation?

Tip: When solving second-order differential equations, always check if one solution is already given. It can greatly simplify finding the second solution through reduction of order!