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The problem in the image involves a differential equation of the form:

\[
\frac{d^4 y}{dx^4} - \frac{d^3 y}{dx^3} = 0
\]

with the given initial conditions:

\[
y = \frac{dy}{dt} = 2, \quad \frac{d^2 y}{dx^2} = \frac{d^3 y}{dx^3} = 1
\]

The goal seems to be solving the differential equation at \( x = 0 \), and you are asked to substitute \( y = t \), solve for \( y \) in terms of \( t \), and then replace \( \frac{t}{s} \).

Here's a breakdown of the problem-solving steps:

1. **Analyze the differential equation:**
   The equation \( \frac{d^4 y}{dx^4} - \frac{d^3 y}{dx^3} = 0 \) is a linear homogeneous differential equation, which can be simplified. The characteristic equation corresponding to this would be:

   \[
   r^4 - r^3 = 0 \quad \Rightarrow \quad r^3(r - 1) = 0
   \]

   This gives roots \( r = 0 \) (multiplicity 3) and \( r = 1 \).

2. **General solution:**
   The general solution of the equation is based on the characteristic roots, so:

   \[
   y(x) = C_1 + C_2 x + C_3 x^2 + C_4 e^x
   \]

3. **Apply the initial conditions:**

   Given:
   - \( y(0) = 2 \)
   - \( \frac{dy}{dx}\bigg|_{x=0} = 2 \)
   - \( \frac{d^2 y}{dx^2}\bigg|_{x=0} = 1 \)
   - \( \frac{d^3 y}{dx^3}\bigg|_{x=0} = 1 \)

   From these conditions, we can substitute into the general solution and its derivatives to solve for \( C_1, C_2, C_3, \) and \( C_4 \).

4. **Solve for constants:**
   Using the conditions one by one:
   - \( y(0) = C_1 + C_4 = 2 \) → \( C_1 + C_4 = 2 \)
   - \( \frac{dy}{dx} = C_2 + 2 C_3 x + C_4 e^x \), so \( \frac{dy}{dx}\bigg|_{x=0} = C_2 + C_4 = 2 \)
   - \( \frac{d^2 y}{dx^2} = 2C_3 + C_4 e^x \), so \( \frac{d^2 y}{dx^2}\bigg|_{x=0} = 2C_3 + C_4 = 1 \)
   - \( \frac{d^3 y}{dx^3} = C_4 e^x \), so \( \frac{d^3 y}{dx^3}\bigg|_{x=0} = C_4 = 1 \)

   Solving these gives:
   - \( C_4 = 1 \)
   - \( C_1 = 1 \)
   - \( C_2 = 1 \)
   - \( C_3 = 0 \)

   So, the solution becomes:

   \[
   y(x) = 1 + x + e^x
   \]

5. **Substitution of \( y = t \):**

   If you are asked to substitute \( y = t \), then the equation becomes \( t = 1 + x + e^x \), and you can solve for \( x \) in terms of \( t \).

Do you want more details or have any questions?

### Related Questions:
1. What are the methods to solve homogeneous differential equations?
2. How do you apply initial conditions to find specific solutions?
3. What is the characteristic equation of a linear differential equation?
4. Can you explain the concept of multiplicity in the roots of characteristic equations?
5. How does the substitution of variables affect the solution process?

### Tip:
For solving higher-order differential equations, always start by finding the characteristic equation, which simplifies the process of finding the general solution.

Solve the differential equation \( \frac{d^4 y}{dx^4} - \frac{d^3 y}{dx^3} = 0 \) with initial conditions \( y = 2 \), \( \frac{dy}{dx} = 2 \), \( \frac{d^2 y}{dx^2} = 1 \), \( \frac{d^3 y}{dx^3} = 1 \), and find the solution at \( x = 0 \), replacing \( y = t \).

Learn how to solve a fourth-order differential equation with given initial conditions. This detailed solution explains key concepts like characteristic equations, solving for constants, and applying initial conditions for a general solution. Suitable for advanced undergraduate mathematics courses.

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