We are given the second-order linear differential equation:

$x^2 y'' + 7x y' + 9y = 0$ with the initial conditions $y(1) = 1$ and $y'(1) = 0$.

This is a Cauchy-Euler equation (or Euler-Cauchy equation), which has the general form:

$x^2 y'' + b x y' + c y = 0$

To solve this equation, we assume a solution of the form:

$y = x^r$

Step 1: Compute the derivatives

For $y = x^r$, the first and second derivatives are:

$y' = r x^{r-1}$ $y'' = r(r-1) x^{r-2}$

Step 2: Substitute into the differential equation

Substitute these expressions for $y$, $y'$, and $y''$ into the given differential equation:

$x^2 \cdot r(r-1) x^{r-2} + 7x \cdot r x^{r-1} + 9x^r = 0$

Simplify each term:

$r(r-1) x^r + 7r x^r + 9x^r = 0$

Factor out $x^r$ (which is nonzero for $x \neq 0$):

$x^r \left( r(r-1) + 7r + 9 \right) = 0$

Thus, the characteristic equation is:

$r(r-1) + 7r + 9 = 0$

Step 3: Solve the characteristic equation

Expand and simplify:

$r^2 - r + 7r + 9 = 0$ $r^2 + 6r + 9 = 0$

Factor the quadratic equation:

$(r + 3)^2 = 0$

So, the repeated root is:

$r = -3$

Step 4: General solution

Since we have a repeated root $r = -3$, the general solution to the differential equation is:

$y(x) = (C_1 + C_2 \ln(x)) x^{-3}$

Step 5: Apply initial conditions

Now, apply the initial conditions to find $C_1$ and $C_2$.

Condition 1: $y(1) = 1$

Substitute $x = 1$ and $y(1) = 1$:

$1 = (C_1 + C_2 \ln(1)) 1^{-3}$ $1 = C_1$ Thus, $C_1 = 1$.

Condition 2: $y'(1) = 0$

First, compute the derivative of $y(x)$:

$y'(x) = \frac{d}{dx} \left( (C_1 + C_2 \ln(x)) x^{-3} \right)$ Use the product rule:

$y'(x) = x^{-3} \frac{d}{dx} \left( C_1 + C_2 \ln(x) \right) + \left( C_1 + C_2 \ln(x) \right) \frac{d}{dx} \left( x^{-3} \right)$

$y'(x) = x^{-3} \left( 0 + \frac{C_2}{x} \right) + \left( C_1 + C_2 \ln(x) \right) (-3x^{-4})$ $y'(x) = C_2 x^{-4} - 3 \left( C_1 + C_2 \ln(x) \right) x^{-4}$ $y'(x) = \frac{C_2 - 3C_1 - 3C_2 \ln(x)}{x^4}$

Now, apply the initial condition $y'(1) = 0$:

$0 = \frac{C_2 - 3C_1 - 3C_2 \ln(1)}{1^4}$ $0 = C_2 - 3C_1$ Substitute $C_1 = 1$:

$0 = C_2 - 3$ Thus, $C_2 = 3$.

Step 6: Final solution

The solution to the initial value problem is:

$y(x) = (1 + 3 \ln(x)) x^{-3}$

Summary

The solution to the differential equation is:

$y(x) = (1 + 3 \ln(x)) x^{-3}$

Would you like to see further steps or clarifications? Here are some related questions for practice:

1. What happens if the characteristic equation has two distinct real roots?
2. How would the solution change if we had a nonzero constant term in the differential equation?
3. What is the general solution when the differential equation has complex roots?
4. How does the solution process change for higher-order Euler equations?
5. What other methods can be used to solve second-order linear differential equations?

Tip: Always double-check the initial conditions after solving the differential equation to ensure they are correctly applied.