Chapter 7: Problem 25
In each of Exercises \(17-28,\) solve the given initial value problem. $$ \frac{d y}{d x}+2 y=4 x \quad y(0)=3 $$
Short Answer
Expert verified
The solution to the initial value problem is \( y = 2x - 2 + 5e^{-2x} \).
Step by step solution
01
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.
02
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: \[ \mu(x) = e^{\int P(x) \, dx} = e^{\int 2 \, dx} = e^{2x}. \]
03
Multiply the Equation by the Integrating Factor
Multiply the entire differential equation by the integrating factor \( e^{2x} \):\[e^{2x} \frac{d y}{d x} + 2e^{2x} y = 4x e^{2x}. \]
04
Recognize as an Exact Differential
The left-hand side can now be expressed as the derivative of a product:\[ \frac{d}{dx}(e^{2x} y) = 4x e^{2x}. \]
05
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 \( dv = e^{2x} \, dx \).
06
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:\[ 4x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \cdot 4 \, dx = 2x e^{2x} - 2 e^{2x}. \]
07
Combine the Results and Solve for y
Rewriting, we have:\[ e^{2x} y = 2x e^{2x} - 2 e^{2x} + C. \]Solve for \( y \):\[ y = 2x - 2 + Ce^{-2x}. \]
08
Apply the Initial Condition
Use the initial condition \( y(0) = 3 \):\[ 3 = 2(0) - 2 + C \Rightarrow C = 5. \]
09
Write the General Solution
Substitute \( C \) back into the equation for \( y \):\[ y = 2x - 2 + 5e^{-2x}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
In solving differential equations, addressing initial value problems is crucial for obtaining useful solutions that fulfill real-world scenarios.
- 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.
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:
This approach is particularly useful because it gives a systematic way of reducing a seemingly complex differential equation to a solvable form.
- Identify \(P(x)\) from the differential equation. Here, \(P(x) = 2\).
- Calculate the integrating factor \(\mu(x)\) using the formula \(e^{\int P(x) \, dx}\).
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 \(\int 4x e^{2x} \, dx\):
\[\int u \, dv = uv - \int v \, du\]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 \(v = \frac{1}{2}e^{2x}\).
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:
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.
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 \(\frac{d}{dx}(e^{2x} y)\).
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.