Find y as a function of x if x^2y''-3xy'-12y=0 y(1)=-3,y'(1)=-5

The given differential equation is:

\[
x^2 y'' - 3x y' - 12y = 0
\]

This is a second-order linear homogeneous differential equation with variable coefficients. To solve it, we will try a solution of the form \( y(x) = x^r \), where \( r \) is a constant to be determined.

### Step 1: Assume \( y(x) = x^r \)

- First derivative: \( y'(x) = r x^{r-1} \)
- Second derivative: \( y''(x) = r(r-1) x^{r-2} \)

### Step 2: Substitute into the differential equation

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

\[
x^2 \cdot r(r-1) x^{r-2} - 3x \cdot r x^{r-1} - 12 x^r = 0
\]

Simplify each term:

\[
r(r-1) x^r - 3r x^r - 12 x^r = 0
\]

Factor out \( x^r \):

\[
x^r \left[ r(r-1) - 3r - 12 \right] = 0
\]

Since \( x^r \neq 0 \), we have:

\[
r(r-1) - 3r - 12 = 0
\]

Simplify the quadratic equation:

\[
r^2 - r - 3r - 12 = 0
\]

\[
r^2 - 4r - 12 = 0
\]

### Step 3: Solve the characteristic equation

Solve the quadratic equation:

\[
r^2 - 4r - 12 = 0
\]

Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), and \( c = -12 \):

\[
r = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-12)}}{2(1)}
\]

\[
r = \frac{4 \pm \sqrt{16 + 48}}{2}
\]

\[
r = \frac{4 \pm \sqrt{64}}{2}
\]

\[
r = \frac{4 \pm 8}{2}
\]

Thus, the two roots are:

\[
r_1 = \frac{4 + 8}{2} = 6, \quad r_2 = \frac{4 - 8}{2} = -2
\]

### Step 4: General solution

The general solution to the differential equation is:

\[
y(x) = C_1 x^6 + C_2 x^{-2}
\]

### Step 5: Apply initial conditions

Now, use the initial conditions \( y(1) = -3 \) and \( y'(1) = -5 \) to find \( C_1 \) and \( C_2 \).

1. **Using \( y(1) = -3 \):**

\[
y(1) = C_1(1)^6 + C_2(1)^{-2} = C_1 + C_2 = -3
\]

Thus, we have the first equation:

\[
C_1 + C_2 = -3 \quad \text{(Equation 1)}
\]

2. **Using \( y'(1) = -5 \):**

First, find \( y'(x) \):

\[
y'(x) = 6C_1 x^5 - 2C_2 x^{-3}
\]

Substitute \( x = 1 \) and \( y'(1) = -5 \):

\[
y'(1) = 6C_1(1)^5 - 2C_2(1)^{-3} = 6C_1 - 2C_2 = -5
\]

Thus, we have the second equation:

\[
6C_1 - 2C_2 = -5 \quad \text{(Equation 2)}
\]

### Step 6: Solve the system of equations

We have the system of equations:

1. \( C_1 + C_2 = -3 \)
2. \( 6C_1 - 2C_2 = -5 \)

Solve Equation 1 for \( C_2 \):

\[
C_2 = -3 - C_1
\]

Substitute into Equation 2:

\[
6C_1 - 2(-3 - C_1) = -5
\]

Simplify:

\[
6C_1 + 6 + 2C_1 = -5
\]

\[
8C_1 = -11
\]

\[
C_1 = -\frac{11}{8}
\]

Substitute \( C_1 = -\frac{11}{8} \) into Equation 1:

\[
-\frac{11}{8} + C_2 = -3
\]

\[
C_2 = -3 + \frac{11}{8} = -\frac{24}{8} + \frac{11}{8} = -\frac{13}{8}
\]

### Step 7: Final solution

Thus, the solution to the differential equation is:

\[
y(x) = -\frac{11}{8} x^6 - \frac{13}{8} x^{-2}
\]

Would you like any additional details or further explanation?

Here are 5 related questions:
1. How would you solve a non-homogeneous differential equation of this form?
2. Can you solve the same equation using a power series method?
3. How do initial conditions influence the solution of a differential equation?
4. What is the significance of the characteristic equation in solving linear ODEs?
5. How can you apply this method to systems of differential equations?

**Tip:** When solving second-order differential equations, always check if a trial solution like \( x^r \) works for simpler cases before considering more complex methods.

Find y as a function of x if x^2y''-3xy'-12y=0, y(1)=-3, y'(1)=-5

This page provides a step-by-step solution for solving the differential equation x^2y'' - 3xy' - 12y = 0 with initial conditions y(1) = -3 and y'(1) = -5. It includes the use of characteristic equations, the quadratic formula, and solving for arbitrary constants.

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