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Magazine Chapter 4: Problem 36
36\. Show that the differential equation $$ \frac{d y}{d t}=a y+b, y(0)=y_{0} $$ has solution $$ y=\left(y_{0}+\frac{b}{a}\right) e^{a t}-\frac{b}{a} $$ Assume that \(a \neq 0\).
Short Answer
Expert verified
The solution to the differential equation is \(y = \left(y_{0} + \frac{b}{a}\right) e^{a t} - \frac{b}{a}\).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation \(\frac{dy}{dt} = ay + b\) is a first-order linear differential equation. This type of equation has the form \(\frac{dy}{dt} + P(t)y = Q(t)\). Here, \(P(t) = -a\) and \(Q(t) = b\).
02
Integrating Factor Method
To solve the differential equation, we use the integrating factor method. The integrating factor is calculated as \(e^{\int P(t) dt} = e^{-at}\).
03
Multiply by the Integrating Factor
Multiply the entire differential equation by the integrating factor \(e^{-at}\):\[e^{-at}\frac{dy}{dt} - ae^{-at}y = be^{-at}\]
04
Express the Left Side as a Derivative
Notice that the left side of the equation can be written as the derivative of a product due to the integrating factor:\[\frac{d}{dt}(e^{-at}y) = be^{-at}\]
05
Integrate Both Sides
Integrate both sides with respect to \(t\):\[\int \frac{d}{dt}(e^{-at}y) dt = \int be^{-at} dt\]This gives:\[e^{-at}y = -\frac{b}{a} e^{-at} + C\]
06
Solve for \(y\)
Multiply both sides by \(e^{at}\) to solve for \(y\):\[y = -\frac{b}{a} + Ce^{at}\]
07
Apply Initial Condition
Apply the initial condition \(y(0) = y_0\) to find \(C\):\[y(0) = -\frac{b}{a} + C = y_0\]Solving for \(C\), we get:\[C = y_0 + \frac{b}{a}\]
08
Write the Particular Solution
Substitute \(C\) back into the equation for \(y\):\[y = \left( y_0 + \frac{b}{a} \right) e^{at} - \frac{b}{a}\]
09
Conclusion
Thus, we have shown the differential equation \(\frac{dy}{dt} = ay + b\) with initial condition \(y(0) = y_0\) has the solution:\[y = \left( y_0 + \frac{b}{a} \right) e^{at} - \frac{b}{a}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor Method
When dealing with first-order linear differential equations, the integrating factor method becomes a powerful tool. This method is specifically designed to manage equations of the form \( \frac{dy}{dt} + P(t)y = Q(t) \). Here, \( P(t) \) and \( Q(t) \) are functions of \( t \).
- First, identify \( P(t) \) from the equation. In our case, \( P(t) = -a \) and \( Q(t) = b \).
- The integrating factor, \( \mu(t) \), is determined by calculating \( e^{\int P(t) dt} \). For our problem, it simplifies to \( e^{-at} \).
- Once the integrating factor is known, multiply the entire differential equation by this factor. This process simplifies the left side of the equation into a derivative of a product.
Initial Value Problem
An initial value problem in the context of differential equations specifies conditions that the solution must satisfy initially. This usually takes the form of \( y(t_0) = y_0 \), where \( t_0 \) is the starting point and \( y_0 \) is the initial value of the solution. Applying this initial condition is crucial for finding a specific solution to a differential equation.
- In our exercise, the initial condition given is \( y(0) = y_0 \).
- This helps in determining the constant \( C \) after integrating the equation.
Differential Equation Solution
Solving the differential equation \( \frac{dy}{dt} = ay + b \) involves combining the integrating factor method with the initial value condition. The steps to arrive at the solution \( y = \left(y_0 + \frac{b}{a}\right)e^{at} - \frac{b}{a} \) are as follows:First, by applying the integrating factor \( e^{-at} \), the differential equation can be rewritten in terms of the derivative of a product \( \frac{d}{dt}(e^{-at}y) = be^{-at} \).
- Integrate both sides with respect to \( t \) to find the integrated equation.
- This results in \( e^{-at}y = -\frac{b}{a} e^{-at} + C \).
- To isolate \( y \), multiply through by \( e^{at} \), leading to \( y = -\frac{b}{a} + Ce^{at} \).
- Finally, use the initial condition \( y(0) = y_0 \) to solve for \( C \), giving \( C = y_0 + \frac{b}{a} \).
