Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 \(a \neq 0\).

Step-by-Step Solution

Answer

The solution to the differential equation is \(y = \left(y_{0} + \frac{b}{a}\right) e^{a t} - \frac{b}{a}\).

Step 1: 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 \(Q(t) = b\).

Step 2: Integrating Factor Method

To solve the differential equation, we use the integrating factor method. The integrating factor is calculated as \(e^{\int P(t) dt} = e^{-at}\).

Step 3: Multiply by the Integrating Factor

Multiply the entire differential equation by the integrating factor \(e^{-at}\):\[e^{-at}\frac{dy}{dt} - ae^{-at}y = be^{-at}\]

Step 4: 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:\[\frac{d}{dt}(e^{-at}y) = be^{-at}\]

Step 5: Integrate Both Sides

Integrate both sides with respect to \(t\):\[\int \frac{d}{dt}(e^{-at}y) dt = \int be^{-at} dt\]This gives:\[e^{-at}y = -\frac{b}{a} e^{-at} + C\]

Step 6: Solve for \(y\)

Multiply both sides by \(e^{at}\) to solve for \(y\):\[y = -\frac{b}{a} + Ce^{at}\]

Step 7: 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:\[C = y_0 + \frac{b}{a}\]

Step 8: Write the Particular Solution

Substitute \(C\) back into the equation for \(y\):\[y = \left( y_0 + \frac{b}{a} \right) e^{at} - \frac{b}{a}\]

Step 9: Conclusion

Thus, we have shown the differential equation \(\frac{dy}{dt} = ay + b\) with initial condition \(y(0) = y_0\) has the solution:\[y = \left( y_0 + \frac{b}{a} \right) e^{at} - \frac{b}{a}\]

Key Concepts

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.

This method transforms a potentially complicated differential equation into a solvable format, making the path to the solution much clearer.

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.

Without the initial condition, the solution would include an arbitrary constant, resulting in a family of solutions rather than a single one.

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 \( C = y_0 + \frac{b}{a} \).

Substitute \( C \) back into the equation to obtain the particular solution satisfying the initial condition. This ensures that the solution not only fits the equation but also the initial value specified.