Question

Solve the following initial-value problems by using integrating factors. $$ y^{\prime}+y=x, y(0)=3 $$

Short answer

The solution is \( y = x - 1 + 4e^{-x} \).

Step-by-step solution

Identify the Standard Form and Variables

The differential equation given is \( y' + y = x \). This equation is in the standard form of a linear first-order ordinary differential equation: \( y' + P(x) y = Q(x) \). Here, \( P(x) = 1 \) and \( Q(x) = x \).

Compute the Integrating Factor

The integrating factor \( \mu(x) \) is given by \( e^{\int P(x) \; dx} \). Since \( P(x) = 1 \), we have:\[\mu(x) = e^{\int 1 \; dx} = e^{x}.\]

Multiply Entire Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^x \):\[e^x y' + e^x y = e^x x.\]

Recognize the Left-Hand Side as a Derivative

Notice that the left-hand side is the derivative of \( e^x y \). Therefore, the equation becomes:\[\frac{d}{dx}(e^x y) = e^x x.\]

Integrate Both Sides

Integrate both sides with respect to \( x \):\[\int \frac{d}{dx}(e^x y) \; dx = \int e^x x \; dx.\]The left-hand side simplifies directly to \( e^x y \). The right-hand side can be integrated using integration by parts.

Use Integration by Parts on the Right-Hand Side

Let \( u = x \) and \( dv = e^x dx \). Then \( du = dx \) and \( v = e^x \). By integration by parts, we have:\[\int e^x x \; dx = x e^x - \int e^x \; dx = x e^x - e^x.\]

Solve for the Function \( y \)

After integration, the equation becomes:\[e^x y = x e^x - e^x + C,\]where \( C \) is the constant of integration. Solve for \( y \) by dividing through by \( e^x \):\[y = x - 1 + Ce^{-x}.\]

Use Initial Condition to Solve for \( C \)

Apply the initial condition \( y(0) = 3 \):\[3 = 0 - 1 + C e^{0}3 = -1 + C \times 1.\]Solve for \( C \):\[C = 4.\]

Write the Final Solution

Substitute \( C = 4 \) back into the equation for \( y \):\[y = x - 1 + 4 e^{-x}.\]

Key concepts

Linear First-Order Differential Equation

The core of solving a linear first-order differential equation lies in understanding its standard form. These equations can be expressed as:\[ y' + P(x)y = Q(x) \]Here, the notation \( y' \) represents the derivative of \( y \) with respect to \( x \). \( P(x) \) and \( Q(x) \) are functions of \( x \). The challenge is to solve for \( y \), which depends on \( x \).

To address such equations, we use an integrating factor. This factor, \( \mu(x) \), typically simplifies the process by allowing the equation to be expressed as a derivative of a product:- Identify the equation's form and values of \( P(x) \) and \( Q(x) \).- Calculate the integrating factor \( \mu(x) = e^{\int P(x) \, dx} \).

Once these steps are completed, multiply every component of the equation by \( \mu(x) \). This will transform the left side into a derivative of a product, easing the integration process. This simplification helps in systematically obtaining the solution.

Initial-Value Problem

An initial-value problem is a type of differential equation that comes with an extra piece of information: the initial condition. This condition specifies the value of the function at a particular point, often used to find the constant of integration in the solution.

In our exercise, the initial condition is given as \( y(0) = 3 \). This tells us that when \( x = 0 \), \( y \) equals 3. Such a condition is critical because:- It allows us to uniquely determine the constant \( C \).- It turns the general solution into a specific solution applicable to a particular scenario.

After obtaining the general form of the solution, substitute the initial conditions into the formula. Solve the resulting equation to find the specific value of \( C \). This way, the solution not only satisfies the differential equation but also the initial conditions.

Integration by Parts

Integration by parts is a technique derived from the product rule for differentiation. It is crucial for solving integrals that involve products of functions. The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]Here, \( u \) and \( dv \) are chosen parts of the integrand, and you derive \( du \) and \( v \) through differentiation and integration, respectively.

During our exercise, integration by parts was used to integrate \( \int e^x x \, dx \). We chose:- \( u = x \) \( \rightarrow \) \( du = dx \)- \( dv = e^x \, dx \) \( \rightarrow \) \( v = e^x \)

Substituting into the integration by parts formula gives:\[ \int e^x x \, dx = xe^x - \int e^x \, dx \]The second integral simplifies to \( e^x \), leading to the result \( xe^x - e^x + C \). This technique is especially handy when direct integration is challenging due to the function's complexity.