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Magazine Chapter 9: Problem 15
Solve the initial value problems. \(\frac{d y}{d t}+2 y=3, \quad y(0)=1\)
Short Answer
Expert verified
The solution is \( y = \frac{3}{2} - \frac{1}{2}e^{-2t} \).
Step by step solution
01
Identify the Type of Differential Equation
The equation \( \frac{d y}{d t}+2 y=3 \) is a first-order linear ordinary differential equation. The standard form is \( \frac{d y}{d t} + P(t)y = Q(t) \) where \( P(t) = 2 \) and \( Q(t) = 3 \).
02
Calculate the Integrating Factor
The integrating factor \( \mu(t) \) is given by \( e^{\int P(t) \, dt} = e^{\int 2 \, dt} = e^{2t} \).
03
Multiply the Entire Equation by the Integrating Factor
Multiply both sides of the equation by the integrating factor \( e^{2t} \) to get: \( e^{2t} \frac{d y}{d t} + 2 e^{2t} y = 3 e^{2t} \). This simplifies to \( \frac{d}{dt}(e^{2t}y) = 3 e^{2t} \).
04
Integrate Both Sides with Respect to t
Integrate both sides with respect to \( t \):\( \int \frac{d}{dt}(e^{2t}y) \, dt = \int 3 e^{2t} \, dt \). This gives us \( e^{2t}y = \frac{3}{2} e^{2t} + C \), where \( C \) is a constant.
05
Solve for the Particular Solution Using the Initial Condition
Use the initial condition \( y(0) = 1 \) to find \( C \): \( e^{0} \cdot 1 = \frac{3}{2} \cdot e^{0} + C \). So, \( 1 = \frac{3}{2} + C \) which implies \( C = 1 - \frac{3}{2} = -\frac{1}{2} \).
06
Write the General Solution
Substitute \( C = -\frac{1}{2} \) back into the equation \( e^{2t}y = \frac{3}{2} e^{2t} - \frac{1}{2} \): \( y = \frac{3}{2} - \frac{1}{2}e^{-2t} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
When faced with a first-order linear differential equation such as \( \frac{dy}{dt} + P(t)y = Q(t) \), an integrating factor comes to the rescue. It is a tool that simplifies solving these equations. The integrating factor, denoted \( \mu(t) \), is computed using the formula \( (t) = e^{\int P(t) \, dt} \).
For our equation, where \( P(t) = 2 \), the integrating factor becomes \( e^{2t} \). By multiplying the entire differential equation by \( e^{2t} \), the structure of the equation is transformed. This transformation neatens the left-hand side, turning it into the derivative of a product, specifically \( \frac{d}{dt}(e^{2t}y) \).
After applying the integrating factor and simplifying the equation, we can easily integrate both sides with respect to \( t \). This process allows us to find a more straightforward expression for \( y(t) \).
For our equation, where \( P(t) = 2 \), the integrating factor becomes \( e^{2t} \). By multiplying the entire differential equation by \( e^{2t} \), the structure of the equation is transformed. This transformation neatens the left-hand side, turning it into the derivative of a product, specifically \( \frac{d}{dt}(e^{2t}y) \).
After applying the integrating factor and simplifying the equation, we can easily integrate both sides with respect to \( t \). This process allows us to find a more straightforward expression for \( y(t) \).
- Identify the function \( P(t) \).
- Calculate the integrating factor \( \mu(t) = e^{\int P(t) \, dt} \).
- Multiply the entire differential equation by \( \mu(t) \).
Initial value problem
An initial value problem (IVP) means we solve a differential equation with a condition imposed on the solution at a specific point. This condition is crucial because it helps us determine the constant of integration, allowing us to find exactly one solution instead of a family of solutions.
In the equation \( \frac{dy}{dt} + 2y = 3 \), the initial condition given is \( y(0) = 1 \). This specifies that when \( t = 0 \), the value of \( y \) must be 1. After finding the general solution to the differential equation, we use this condition to find the particular value of the constant \( C \).
In the equation \( \frac{dy}{dt} + 2y = 3 \), the initial condition given is \( y(0) = 1 \). This specifies that when \( t = 0 \), the value of \( y \) must be 1. After finding the general solution to the differential equation, we use this condition to find the particular value of the constant \( C \).
- Identify given initial condition \( y(t_0) = y_0 \).
- Integrate and find the general solution.
- Substitute the initial condition to solve for the constant.
Particular solution
The particular solution to a differential equation is the version of the general solution that meets the specific initial conditions provided by an IVP.
Initially, solving the differential equation gives us a general solution that includes a constant \( C \), representing a family of possible solutions. Here, the integration of \( \int 3e^{2t} \, dt \) gives us \( \frac{3}{2}e^{2t} + C \).
By applying the initial condition \( y(0) = 1 \), we substitute \( t = 0 \) and \( y = 1 \) into the expression, finding that \( C = -\frac{1}{2} \).
Thus, the particular solution becomes: \( y(t) = \frac{3}{2} - \frac{1}{2}e^{-2t} \). This solution not only satisfies the differential equation but also aligns with the initial value, capturing the exact behavior of the system at \( t = 0 \).
Initially, solving the differential equation gives us a general solution that includes a constant \( C \), representing a family of possible solutions. Here, the integration of \( \int 3e^{2t} \, dt \) gives us \( \frac{3}{2}e^{2t} + C \).
By applying the initial condition \( y(0) = 1 \), we substitute \( t = 0 \) and \( y = 1 \) into the expression, finding that \( C = -\frac{1}{2} \).
Thus, the particular solution becomes: \( y(t) = \frac{3}{2} - \frac{1}{2}e^{-2t} \). This solution not only satisfies the differential equation but also aligns with the initial value, capturing the exact behavior of the system at \( t = 0 \).
- Solve for \( C \) using initial conditions.
- Ensure the solution satisfies both the differential equation and the initial values.
