Question

Solve the following initial-value problems with the initial condition \(y_{0}=0\) and graph the solution.\(\frac{d y}{d t}=y-1\)

Short answer

The solution is \( y(t) = -e^t + 1 \) with an initial condition \( y_0 = 0 \).

Step-by-step solution

Identify the equation type

The given differential equation is \( \frac{d y}{d t} = y - 1 \), which is a first-order linear ordinary differential equation (ODE).

Solve the homogeneous equation

Start by solving the associated homogeneous equation: \( \frac{d y}{d t} = y \). This is separable, and integrating both sides yields: \( \ln|y| = t + C \), where \( C \) is the integration constant. Therefore, the solution to the homogeneous equation is \( y_h(t) = Ce^t \).

Solve the particular solution

To solve the non-homogeneous part, assume a constant particular solution \( y_p(t) = C_p \). Substitute \( y = C_p \) into the differential equation \( \frac{d y}{d t} = y - 1 \) to get \( 0 = C_p - 1 \). Thus, \( C_p = 1 \), so the particular solution is \( y_p(t) = 1 \).

General solution

Combine the homogeneous and particular solutions to get the general solution: \( y(t) = y_h(t) + y_p(t) = Ce^t + 1 \).

Apply the initial condition

Use the initial condition \( y(0) = 0 \) to find \( C \). Substitute \( t = 0 \) and \( y = 0 \) into the general solution: \( 0 = Ce^0 + 1 \) which simplifies to \( C = -1 \). Thus, the solution with the initial condition is \( y(t) = -e^t + 1 \).

Graph the solution

Plot the solution \( y(t) = -e^t + 1 \) over a suitable range of \( t \). The graph is an exponential curve shifted downward, starting at \( y_0 = 0 \) when \( t = 0 \), and decreases as \( t \) increases.

Key concepts

Initial Value Problem

An initial value problem in the context of differential equations is a problem where we seek a specific solution that not only satisfies the differential equation but also adheres to a given initial condition. In our exercise, the initial condition provided is \( y(0) = 0 \).
This means that at the time \( t = 0 \), the value of the function \( y(t) \) should be exactly zero. These initial conditions are essential as they help in determining the exact constant, \( C \), in our general solution.
This allows us to fine-tune our solution to fit the unique scenario described by the initial condition.
Solving an initial value problem involves first finding the general solution of the differential equation and then applying the initial condition to find any unknown constants. This specificity turns a broad solution into one that is tailor-made for the problem at hand.

General Solution

The general solution of a differential equation encompasses all possible solutions without the imposition of a specific initial condition.
In our exercise, the equation \( \frac{d y}{d t} = y - 1 \) is a first-order linear ordinary differential equation.
The solution process involves solving both the homogeneous part and a particular solution.

• The homogeneous solution for \( \frac{d y}{d t} = y \) is found to be \( y_h(t) = Ce^t \).
• The particular solution is a constant \( y_p(t) = 1 \), since substituting \( y = 1 \) makes the derivative zero.

The general solution thus becomes \( y(t) = y_h(t) + y_p(t) = Ce^t + 1 \).
It includes the constant \( C \), which will be determined using the initial value given.

Homogeneous Equation

When dealing with first-order linear ODEs, the homogeneous equation is derived by setting the right side to zero.
In our case, the homogeneous form of \( \frac{d y}{d t} = y - 1 \) results in \( \frac{d y}{d t} = y \).
Solving this provides us with a crucial component of our general solution.
The method used for solving it is known as separation of variables, a technique where we integrate both sides after rearranging terms.

• This yields \( \ln|y| = t + C \), resulting in the solution \( y_h(t) = Ce^t \).

This is an exponential function determined by \( C \), an arbitrary constant.
The simplicity of exponential solutions makes this method very powerful in linear ODEs.

Particular Solution

A particular solution in the context of differential equations is a solution that accounts for the non-homogeneous part of the equation.
It does not contain arbitrary constants and solves the differential equation outright.
For our differential equation \( \frac{d y}{d t} = y - 1 \), we assume a constant particular solution \( y_p(t) = C_p \).

• Substituting this into the equation gives \( 0 = C_p - 1 \), solving which results in \( C_p = 1 \).

Thus, the particular solution is \( y_p(t) = 1 \).
This step complements the homogeneous solution we found earlier, allowing us to construct the full general solution.
By piecing together the homogeneous and particular solutions, the general solution becomes \( y(t) = Ce^t + 1 \).
Hence, finding a particular solution is essential to account for non-homogeneous terms in the differential equation.