Question

Find the particular solution to the differential equation \(\frac{d y}{d t}=e^{(t+y)}\) that passes through \((1,0)\), given that \(y=-\ln \left(C-e^{t}\right)\) is a general solution.

Short answer

The particular solution is \( y = -\ln((1 + e) - e^t) \).

Step-by-step solution

Verify General Solution

We are given that the general solution of the differential equation \( \frac{dy}{dt} = e^{(t+y)} \) is \( y = -\ln(C - e^t) \). This satisfies the differential equation and captures all possible solutions parameterized by \( C \).

Identify Initial Condition

The initial condition provided is the point \((t = 1, y = 0)\). We will use this initial condition to find the value of \( C \) specific to the solution that passes through this point.

Substitute Initial Condition and Solve for C

Using the general solution \( y = -\ln(C - e^t) \), substitute \( t = 1 \) and \( y = 0 \):\[ 0 = -\ln(C - e^1) \]\[ \ln(C - e) = 0 \]Take the exponential of both sides to solve for \( C \):\[ C - e = 1 \Rightarrow C = 1 + e \]

Write the Particular Solution

Replace \( C \) in the general solution with the value found: \( C = 1 + e \).The particular solution is:\[ y = -\ln((1 + e) - e^t) \]

Validate the Solution

Verify that the particular solution satisfies both the differential equation and the initial condition by substituting back and checking if the equations hold true.

Key concepts

General Solution

In the context of differential equations, a "general solution" represents a family of solutions that contains all possible solutions of a differential equation. For instance, if we have the differential equation \( \frac{dy}{dt} = e^{(t+y)} \), a general solution to this equation is \( y = -\ln(C - e^t) \).
The general solution is expressed in terms of a parameter \( C \), which allows it to represent a broad set of solutions. This form comes from the integration of the differential equation, considering some constant of integration.

For first-order differential equations, the general solution is particularly useful because it embodies every solution that could exist. By modifying the parameter \( C \), you can adjust the solution to fit different scenarios or constraints, such as specific starting conditions known as initial conditions.

Initial Condition

The concept of an "initial condition" in differential equations is crucial because it helps to pinpoint a specific solution from the family of solutions represented by the general solution. An initial condition gives specific values for the variables in the context of the problem.
In our exercise, the initial condition is given as the point \((t = 1, y = 0)\). This tells us that when \( t = 1 \), the solution \( y \) should take the value 0. Such conditions are necessary to determine the constant \( C \) in the general solution.

• The format for an initial condition is typically a point (like \((t, y)\)) that the solution must pass through.
• By substituting the initial condition into the general solution, we can solve for \( C \), thereby personalizing the solution to fit the exact situation described.

Particular Solution

A "particular solution" refers to a single, unique solution derived from the general solution, which satisfies both the differential equation and the initial conditions, creating a tailor-fit solution.
In our task, using the initial condition \((1,0)\), we determined the parameter \( C \) specific to that condition within the general solution formula \( y = -\ln(C - e^t)\).

Through substituting the initial condition values and solving for \( C \), we arrived at \( C = 1 + e \). This yields the particular solution:
\[ y = -\ln((1 + e) - e^t) \]

• The particular solution fits the boundary of the initial conditions.
• It is unique to the problem constraints and tailored to the "initial snapshot" provided by the initial condition.
• Verifying a particular solution ensures it satisfies both the form of the differential equation and the specified initial conditions, confirming its correctness.