https://file.aiquickdraw.com/mathbot/file/1730950069604rxxd1zzc.jpg

To solve this problem, we need to solve the fourth-order differential equation:

\[
y^{(4)} - 12y^{(3)} + 36y'' = 0
\]

with initial conditions:

\[
y(0) = 2, \quad y'(0) = 16, \quad y''(0) = 36, \quad y^{(3)}(0) = 0
\]

### Step 1: Solve the Differential Equation

The differential equation \( y^{(4)} - 12y^{(3)} + 36y'' = 0 \) suggests a constant-coefficient characteristic equation:
\[
r^4 - 12r^3 + 36r^2 = 0
\]

We can factor this as:
\[
r^2 (r^2 - 12r + 36) = 0
\]

This further factors into:
\[
r^2 (r - 6)^2 = 0
\]

Thus, the roots of the characteristic equation are \( r = 0 \) (with multiplicity 2) and \( r = 6 \) (with multiplicity 2).

### Step 2: Form the General Solution

Given the roots, the general solution for \( y(x) \) is:
\[
y(x) = (C_1 + C_2 x) + (C_3 + C_4 x)e^{6x}
\]

### Step 3: Apply Initial Conditions

Now, we apply the initial conditions to find \( C_1 \), \( C_2 \), \( C_3 \), and \( C_4 \).

1. **Using \( y(0) = 2 \):**
   \[
   y(0) = C_1 + C_3 = 2
   \]

2. **Using \( y'(0) = 16 \):**

   The derivative of \( y(x) \) is:
   \[
   y'(x) = C_2 + (6C_3 + C_4)e^{6x} + 6C_4 x e^{6x}
   \]

   At \( x = 0 \):
   \[
   y'(0) = C_2 + 6C_3 = 16
   \]

3. **Using \( y''(0) = 36 \):**

   The second derivative of \( y(x) \) is:
   \[
   y''(x) = (6C_4 + 36C_3)e^{6x} + 36C_4 x e^{6x}
   \]

   At \( x = 0 \):
   \[
   y''(0) =

Find y as a function of x if y(0) = 2, y'(0) = 16, y''(0) = 36, y'''(0) = 0, and y^(4) - 12y''' + 36y'' = 0.

Learn how to find the function y(x) for a fourth-order differential equation with given initial conditions. This problem involves solving the characteristic equation and applying initial conditions to find the specific solution.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation