Problem 25 - Calculus Single Variable

Question

In each of Exercises \(17-28,\) solve the given initial value problem. $$ \frac{d y}{d x}+2 y=4 x \quad y(0)=3 $$

Step-by-Step Solution

Answer

The solution to the initial value problem is \( y = 2x - 2 + 5e^{-2x} \).

Step 1: Identify the Type of Differential Equation

The given first-order linear differential equation is \( \frac{d y}{d x} + 2y = 4x \) with the initial condition \( y(0) = 3 \). This is a linear first-order differential equation that can be solved using an integrating factor.

Step 2: Calculate the Integrating Factor

The standard form of a differential equation is \( \frac{d y}{d x} + P(x)y = Q(x) \). Here, \( P(x) = 2 \) and \( Q(x) = 4x \). The integrating factor \( \mu(x) \) is calculated as: \[ \mu(x) = e^{\int P(x) \, dx} = e^{\int 2 \, dx} = e^{2x}. \]

Step 3: Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^{2x} \):\[e^{2x} \frac{d y}{d x} + 2e^{2x} y = 4x e^{2x}. \]

Step 4: Recognize as an Exact Differential

The left-hand side can now be expressed as the derivative of a product:\[ \frac{d}{dx}(e^{2x} y) = 4x e^{2x}. \]

Step 5: Integrate Both Sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(e^{2x} y) \, dx = \int 4x e^{2x} \, dx. \]The left side simplifies to \( e^{2x} y \), and the right side can be solved using integration by parts where \( u = 4x \) and \( dv = e^{2x} \, dx \).

Step 6: Solve the Right Side Using Integration by Parts

Set \( u = 4x \) and \( dv = e^{2x} \, dx \). Then \( du = 4 \, dx \) and \( v = \frac{1}{2}e^{2x} \). Integration by parts formula is \( \int u \, dv = uv - \int v \, du \), giving:\[ 4x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \cdot 4 \, dx = 2x e^{2x} - 2 e^{2x}. \]

Step 7: Combine the Results and Solve for y

Rewriting, we have:\[ e^{2x} y = 2x e^{2x} - 2 e^{2x} + C. \]Solve for \( y \):\[ y = 2x - 2 + Ce^{-2x}. \]

Step 8: Apply the Initial Condition

Use the initial condition \( y(0) = 3 \):\[ 3 = 2(0) - 2 + C \Rightarrow C = 5. \]

Step 9: Write the General Solution

Substitute \( C \) back into the equation for \( y \):\[ y = 2x - 2 + 5e^{-2x}. \]

Key Concepts

Initial Value Problem

An initial value problem involves solving a differential equation subject to a given condition, known as an initial condition. The initial condition provides a specific value of the function at a particular point, often helping to determine the constant of integration in the solution.

• The given initial value problem involves the differential equation \(\frac{d y}{d x} + 2y = 4x\).
• The initial condition is \(y(0) = 3\), meaning that when \(x = 0\), the value of \(y\) is 3.

This initial condition simplifies finding a precise solution to the differential equation. It ties the general solution to a specific solution, giving a complete answer rather than an indefinite one where the constant \(C\) is unknown.
In solving differential equations, addressing initial value problems is crucial for obtaining useful solutions that fulfill real-world scenarios.

Integrating Factor

The integrating factor method is a powerful technique used to solve linear first-order differential equations. This method involves multiplying the entire equation by a special function, known as the integrating factor, which transforms it into a form where the left side becomes an exact derivative. In our case, it enables solving:

• Identify \(P(x)\) from the differential equation. Here, \(P(x) = 2\).
• Calculate the integrating factor \(\mu(x)\) using the formula \(e^{\int P(x) \, dx}\).

For the given problem, \(\mu(x) = e^{2x}\). By multiplying the whole differential equation by this integrating factor, the left-hand side transforms, allowing us to use simpler integration methods.
This approach is particularly useful because it gives a systematic way of reducing a seemingly complex differential equation to a solvable form.

Integration by Parts

Integration by parts is an integration technique derived from the product rule of differentiation. When faced with an integral that involves a product of functions, this method becomes invaluable. The principle is based on the formula:
\[\int u \, dv = uv - \int v \, du\]In the original step-by-step solution, we apply this method on the right-hand side of the equation after the integrating factor. For \(\int 4x e^{2x} \, dx\):

• Choose \(u = 4x\) and \(dv = e^{2x} \, dx\).
• Then, \(du = 4 \, dx\) and \(v = \frac{1}{2}e^{2x}\).

Thus, integration by parts simplifies to \(2x e^{2x} - 2 e^{2x}\). Applying this method helps tackle complex integrals by breaking them into simpler parts, making mathematical solutions more accessible.

Exact Differential

An exact differential equation is one where the left side of the equation can be expressed as the derivative of a product of functions. Achieving this form often simplifies solving the differential equation. In our scenario:
After multiplying by the integrating factor, the equation transforms:

• The equation \(e^{2x} \frac{d y}{d x} + 2e^{2x} y = 4x e^{2x}\) alters to \(\frac{d}{dx}(e^{2x} y)\).

An exact differential \(\frac{d}{dx}(e^{2x} y)\) indicates that taking its integral is straightforward, reverting the differentiation step.
Recognizing and utilizing an exact differential eases the integration process, facilitating the solution to differential equations. This method is key to solving linear first-order differential equations efficiently, especially when combined with techniques like integrating factors.