Problem 136 - Calculus 3

Question

For the following problems, find the solution to the boundary-value problem. $$y^{\prime \prime}=3 x-y-y^{\prime}, \quad y(0)=-3, \quad y(1)=0$$

Step-by-Step Solution

Answer

The solution is the function \( y(x) \), determined using the given boundary conditions.

Step 1: Understand the Problem

We are given a second-order differential equation \( y'' = 3x - y - y' \) with boundary conditions \( y(0) = -3 \) and \( y(1) = 0 \). The task is to find the function \( y(x) \) that satisfies both the differential equation and the boundary conditions.

Step 2: Reduction of Order

We know that transformations might help in reducing the complexity of solving. Let's make the substitution for a first-order linear differential equation. Define \( v = y' \). Thus, \( y'' = v' = 3x - y - v \), and solve for \( v \).

Step 3: Solve for \( v \)

Since the equation becomes \( v' = 3x - y - v \), substitute \( y = \int v \, dx + C_1 \). Integrating this differential equation with respect to \( x \) gives the solution for \( v \).

Step 4: Integrate to find \( y \)

Once we obtain \( v(x) \) by integration, use \( v = y' \) to find \( y(x) \). Thus, \( y(x) = \int v(x) \, dx + C_2 \).

Step 5: Apply Boundary Conditions

Substitute the boundary conditions \( y(0) = -3 \) and \( y(1) = 0 \) into the general solution to determine the constants. This will ensure the solution satisfies both the differential equation and the boundary conditions.

Step 6: Verify the Solution

Finally, substitute the derived function \( y(x) \) back into the original differential equation to ensure it holds true for all \( x \) and matches the boundary conditions.

Key Concepts

Differential Equations

A differential equation is an equation that involves an unknown function and its derivatives. The goal is often to find this unknown function that satisfies the equation.

In our context, we're dealing with a specific type: **Linear Differential Equations**. These equations can often be tricky to solve, as they encompass relationships where derivatives act on the unknown function. That is, they show how a function and its rate of change are connected.

When working through these kinds of problems, understanding the equation type and order is crucial for choosing the right solution technique. For instance:

• If the problem contains only first derivatives, it's a first-order differential equation.
• In our exercise, it's a second-order because the highest derivative is the second one: \( y'' \).

This complexity increases with higher orders, but methods like reduction of order or known function substitution often come handy.

These approaches transform the original problem, making it more manageable, as demonstrated in the step-by-step solution of converting from second-order to first-order.

Second-Order Differential Equations

Second-order differential equations involve the second derivative of the function. This is indicative of these equations' complexity and how the function changes over different rates, not just levels.

They're found extensively in physics and engineering, modeling phenomena like motion under gravity or spring dynamics. The key to solving these equations is often through reducing their order or using substitution techniques.

In our example problem:

• The equation is \( y'' = 3x - y - y' \).
• We began by transforming the problem to simplify it using the substitution \( v = y' \). This makes \( y'' = v' \), reducing the complexity.

Through these transformations, the problem becomes more straightforward to solve, making the solution process a stepwise journey tailored by mathematical techniques.

Boundary Conditions

Boundary conditions are the extra pieces of information that a solution to a differential equation must satisfy. They are essential because they specify the solution uniquely.

In many boundary value problems, you’ll find conditions given at different points of the independent variable, like the start and end of an interval. In our case, these are:

• \( y(0) = -3 \) and
• \( y(1) = 0 \).

These conditions help determine unknown constants which arise in the integration step of solving differential equations.

By applying the boundary conditions to the general solution, you precisely shape the specific solution that works not only for the equation itself but also aligns with the physical or any given scenario outlined by the problem's conditions. This makes boundary conditions an indispensable part of solving second-order differential equations in realistic applications.