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Magazine Chapter 19: Problem 3
Find the general solution of \(\frac{\mathrm{d} x}{\mathrm{~d} t}=2 x+4 t\). What is the particular solution which satisfies \(x(1)=2 ?\)
Short Answer
Expert verified
The general solution is \( x(t) = 2t - 1 + Ce^{-2t} \); the particular solution is \( x(t) = 2t - 1 + e^2 e^{-2t} \).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation \( \frac{\mathrm{d} x}{\mathrm{~d} t} = 2x + 4t \) is a first-order linear non-homogeneous differential equation. It can be expressed in the general form \( \frac{\mathrm{d} x}{\mathrm{~d} t} + P(t)x = Q(t) \), where \( P(t) = -2 \) and \( Q(t) = 4t \).
02
Find the Integrating Factor
To solve the differential equation, we first need to calculate the integrating factor, \( \mu(t) \), which is given by \( \mu(t) = e^{\int P(t) \, dt} \). Here, \( P(t) = 2 \), so the integrating factor is \( \mu(t) = e^{\int 2 \, dt} = e^{2t} \).
03
Solve the Differential Equation Using the Integrating Factor
Multiply the entire differential equation by the integrating factor \( e^{2t} \):\[ e^{2t} \frac{\mathrm{d} x}{\mathrm{~d} t} = e^{2t}(2x + 4t) \]The left-hand side becomes the derivative of \( x \cdot e^{2t} \), so:\[ \frac{\mathrm{d}}{\mathrm{~d} t}(x e^{2t}) = 4t e^{2t} \]Integrate both sides with respect to \( t \):\[ x e^{2t} = \int 4t e^{2t} \, dt \]
04
Integrate the Right-Hand Side
Use integration by parts to evaluate \( \int 4t e^{2t} \, dt \). Let \( u = 4t \) and \( dv = e^{2t} \, dt \). Then \( du = 4 \, dt \) and \( v = \frac{1}{2}e^{2t} \).Integration by parts formula: \( \int u \, dv = uv - \int v \, du \).This gives:\[ \int 4t e^{2t} \, dt = 4t \cdot \frac{1}{2} e^{2t} - \int \frac{1}{2}e^{2t} \cdot 4 \, dt \]\[ = 2t e^{2t} - 2 \int e^{2t} \, dt \]\[ = 2t e^{2t} - 2 \cdot \frac{1}{2}e^{2t} = 2t e^{2t} - e^{2t} \]\[ = e^{2t}(2t - 1) \]
05
Solve for x(t)
Equating the integration done above with \( x e^{2t} \), we have:\[ x e^{2t} = e^{2t}(2t - 1) + C \]Where \( C \) is an integration constant. Divide both sides by \( e^{2t} \) to solve for \( x \):\[ x(t) = 2t - 1 + Ce^{-2t} \]This is the general solution of the differential equation.
06
Find the Particular Solution
To find the particular solution that satisfies \( x(1) = 2 \), substitute \( t = 1 \) and \( x = 2 \) into the general solution:\[ 2 = 2(1) - 1 + Ce^{-2(1)} \]\[ 2 = 2 - 1 + Ce^{-2} \]\[ 2 = 1 + Ce^{-2} \]\[ 1 = Ce^{-2} \]\[ C = e^2 \]The particular solution is:\[ x(t) = 2t - 1 + e^2 e^{-2t} \]
07
Final Solution Summary
The general solution to the differential equation is \( x(t) = 2t - 1 + Ce^{-2t} \). The particular solution, satisfying \( x(1) = 2 \), is \( x(t) = 2t - 1 + e^2 e^{-2t} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-order Linear Differential Equations
First-order linear differential equations are a fundamental component of differential equations. They involve an unknown function and its first derivative. The standard form of a first-order linear differential equation is written as:
\[ \frac{\mathrm{d}x}{\mathrm{d}t} + P(t)x = Q(t) \]
Here, \(P(t)\) and \(Q(t)\) are known functions of the variable \(t\), and \(x\) is the function of \(t\) that we need to determine.
Key characteristics of these equations include:
Understanding the structure of these equations is crucial to solving them effectively.
\[ \frac{\mathrm{d}x}{\mathrm{d}t} + P(t)x = Q(t) \]
Here, \(P(t)\) and \(Q(t)\) are known functions of the variable \(t\), and \(x\) is the function of \(t\) that we need to determine.
Key characteristics of these equations include:
- They are linear because the unknown function and its derivative appear to the power of one.
- They can either be homogeneous, where \(Q(t) = 0\), or non-homogeneous, where \(Q(t) eq 0\).
Understanding the structure of these equations is crucial to solving them effectively.
Integrating Factor Method
The integrating factor method is a powerful technique used to solve first-order linear differential equations. This method transforms the differential equation into a form that can be easily integrated.
Here's how it works:
\[ \mu(t) = e^{\int 2 \, dt} = e^{2t} \]
We then multiply the differential equation by \(e^{2t}\), transforming it so that the left side becomes the derivative of \(x \, e^{2t}\). This change dramatically simplifies integrating both sides to solve for \(x(t)\).
Using an integrating factor is key to converting the differential equation into a more manageable form that reveals the solution.
Here's how it works:
- Identify the integrating factor, which is given by \(\mu(t) = e^{\int P(t) \, dt}\).
- Multiply the entire differential equation by this integrating factor.
- The left side of the equation converts into the derivative of a product, simplifying the equation.
\[ \mu(t) = e^{\int 2 \, dt} = e^{2t} \]
We then multiply the differential equation by \(e^{2t}\), transforming it so that the left side becomes the derivative of \(x \, e^{2t}\). This change dramatically simplifies integrating both sides to solve for \(x(t)\).
Using an integrating factor is key to converting the differential equation into a more manageable form that reveals the solution.
Particular Solution
The particular solution of a differential equation is a solution that satisfies both the differential equation and a specific initial condition. The initial condition typically specifies the value of the function at a particular point.
In the exercise, the initial condition given is \(x(1) = 2\). We use this to find a specific value for the constant \(C\) in the general solution:
\[ x(t) = 2t - 1 + Ce^{-2t} \]
Substituting \(t = 1\) and \(x = 2\) into this equation helps us solve for \(C\):
\[ 2 = 2(1) - 1 + Ce^{-2} \]
Solving for \(C\), we find:
\[ 1 = Ce^{-2} \]
\[ C = e^2 \]
Thus, the particular solution, fitting the given condition, is:
\[ x(t) = 2t - 1 + e^2 e^{-2t} \]
This step ensures that the solution is uniquely tailored to the initial condition, providing the exact behavior of the solution at a specific point in time. The particular solution is crucial in many real-world applications where initial conditions are known.
In the exercise, the initial condition given is \(x(1) = 2\). We use this to find a specific value for the constant \(C\) in the general solution:
\[ x(t) = 2t - 1 + Ce^{-2t} \]
Substituting \(t = 1\) and \(x = 2\) into this equation helps us solve for \(C\):
\[ 2 = 2(1) - 1 + Ce^{-2} \]
Solving for \(C\), we find:
\[ 1 = Ce^{-2} \]
\[ C = e^2 \]
Thus, the particular solution, fitting the given condition, is:
\[ x(t) = 2t - 1 + e^2 e^{-2t} \]
This step ensures that the solution is uniquely tailored to the initial condition, providing the exact behavior of the solution at a specific point in time. The particular solution is crucial in many real-world applications where initial conditions are known.
