Question

Solutien of \(d y / d x+y=0\) given \(y(0)=5\) is

Short answer

The solution is \(y = 5e^{-x}\).

Step-by-step solution

Identify the Type of Differential Equation

The given equation \(\frac{dy}{dx} + y = 0\) is a first-order linear homogeneous differential equation. Our main goal is to solve this equation and find the function \(y(x)\).

Rearrange the Equation to Solve for Integrating Factor

For a first-order linear differential equation of the form \(\frac{dy}{dx} + P(x)y = 0\), the integrating factor method can be used. Here, \(P(x) = 1\). First, we need to find the integrating factor \(\mu(x) = e^{\int P(x) \, dx} = e^{\int 1 \, dx} = e^{x}\).

Multiply Through By the Integrating Factor

Multiply the original equation by the integrating factor \(e^x\): \[ e^x \cdot \frac{dy}{dx} + e^x \cdot y = 0 \]This can be recognized as the derivative of \(e^x y\), thus re-writing:\[ \frac{d}{dx}(e^x y) = 0 \]

Integrate Both Sides

Integrate both sides of the equation \(\frac{d}{dx}(e^x y) = 0\):\[ \int \frac{d}{dx}(e^x y) \, dx = \int 0 \, dx \]This yields:\[ e^x y = C \]where \(C\) is the constant of integration.

Solve for y

To solve for \(y\), divide both sides by \(e^x\):\[ y = C e^{-x} \]

Apply the Initial Condition

We use the initial condition \(y(0) = 5\) to find the constant \(C\). Substitute \(x = 0\) and \(y = 5\) into the equation:\[ 5 = C e^{0} \]Thus, \(C = 5\).

Write the Final Solution

Substitute \(C = 5\) back into the expression for \(y\) to get the particular solution:\[ y = 5 e^{-x} \].

Key concepts

Homogeneous Differential Equation

A homogeneous differential equation is one where all terms are a function of the dependent variable and its derivatives. In the case of first-order linear differential equations, it takes the form \( \frac{dy}{dx} + P(x)y = 0 \). Here, both the terms involve \( y \), and none is a function of \( x \) alone, indicating that the equation is homogeneous. Homogeneous equations often have constant or zero solutions. Recognizing the structure of such equations helps in applying specific methods like the integrating factor to solve them.
In our specific exercise, the given equation is \( \frac{dy}{dx} + y = 0 \). This is a textbook example of a homogeneous differential equation where \( P(x) = 1 \). Once identified, we can confidently apply the integrating factor method to find the solution for \( y(x) \).

Integrating Factor

The integrating factor is a clever technique used to solve linear differential equations. Imagine wanting to simplify an equation like \( \frac{dy}{dx} + P(x)y = 0 \). The integrating factor \( \mu(x) \) makes this possible by converting the equation into a form that's easier to solve by integration.
The formula to find the integrating factor is \( \mu(x) = e^{\int P(x) \, dx} \). In our problem, you calculate it as \( e^{\int 1 \, dx} = e^x \).

• Multiply through by the integrating factor: The step converts the differential equation into one that resembles the derivative of a product.
• Recognize and differentiate: The new form is then simply the derivative of \( e^x y \).
• The simplification then allows easy integration of both sides.

By this technique, we've systematically structured the equation to make solving it straightforward.

Initial Condition Method

The initial condition method is crucial for finding particular solutions to differential equations. While the general solution sets the foundation, applying initial conditions personalizes it to fit specific real-world scenarios.
For the general solution derived, \( y = Ce^{-x} \), inserting the initial condition \( y(0) = 5 \) determines the unique constant \( C \). Here's how it works:

• Substitute \( x = 0 \) into the general solution, leading to the equation: \( 5 = Ce^{0} \).
• Since \( e^{0} = 1 \), this simplifies to \( 5 = C \).

The constant \( C \) thus simplifies the problem, allowing you to craft a definitive solution that aligns with the initial condition. This tailored solution captures the behavior of the system specifically for the given situation.

Constant of Integration

The constant of integration, often represented by \( C \), is a fundamental concept when integrating differential equations. It arises because integration, being the inverse of differentiation, can undo the derivative of infinitely many constant functions—all differing by a constant value.
Applying an integrating factor in our problem led to the equation \( e^x y = C \). This constant arises after integrating, indicating that without specific constraints, multiple functions can satisfy the initial differential equation.
Once you apply an initial condition, like \( y(0) = 5 \), this determines \( C \), helping pinpoint the precise solution. In practical problems, constants customize general solutions, tailoring them to specific initial positions or velocities in physical systems. Understanding this idea ensures clarity in how differential equations mirror real-world scenarios.