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.