Question

The general solution of a first-order linear differential equation is \(y(t)=C e^{-10 t}-13 .\) What solution satisfies the initial condition \(y(0)=4 ?\)

Short answer

Question: Find the particular solution of the first-order linear differential equation \(y(t) = Ce^{-10t} - 13\) that satisfies the initial condition \(y(0) = 4\). Answer: \(y(t) = 17e^{-10t} - 13\)

Step-by-step solution

Plug in the initial condition

Substitute \(t=0\) and \(y(0)=4\) into the general solution: \(4 = Ce^{-10(0)} - 13\)

Solve for C

Since \(e^{-10(0)} = e^{0} = 1\), the equation becomes: \(4 = C - 13\) Now, add 13 to both sides and we get: \(C = 17\)

Write the particular solution

Replace C with the calculated value in the general solution: \(y(t) = 17e^{-10t} - 13\) So the particular solution of the given first-order linear differential equation that satisfies the initial condition \(y(0) = 4\) is \(y(t) = 17e^{-10t} - 13\).

Key concepts

Initial Condition

An initial condition is a crucial element in solving differential equations, especially when looking for a specific solution. It refers to the specific value or state of the function at a certain point, usually at the beginning or a particular moment of interest. In the context of solving a differential equation, an initial condition is given as a pair of values, usually for both the function and its independent variable.
In this exercise, our initial condition is provided as \(y(0) = 4\). This tells us the value of the function \(y\) when \(t = 0\). By applying this condition, we find a precise value for any arbitrary constants in the general solution, allowing us to find the particular solution that fits the specified scenario.

• The initial condition acts as a guideline to narrow down infinite solutions to the one of interest.
• It is essential in applications such as physics and engineering where real-world constraints define specific solutions.

Understanding initial conditions helps in pinpointing a unique solution that meets given criteria at a specified starting point.

Particular Solution

The particular solution of a differential equation is what you get after applying the initial condition to the general solution. It is a specific form of the solution that satisfies both the differential equation and the initial conditions given.
To find it, you start with the general solution and substitute the initial conditions to solve for any arbitrary constants. In this exercise, the general solution is \(y(t) = Ce^{-10t} - 13\). By applying the initial condition \(y(0) = 4\), we calculated \(C = 17\).

• Once \(C\) is identified, replace it back into the general solution to obtain the particular solution.
• The particular solution provides a single, definitive function \(y(t)\) that meets the given conditions.

Thus, the particular solution for this problem is \(y(t) = 17e^{-10t} - 13\), which ensures the equation satisfies \(y(0)=4\).

General Solution

The general solution of a first-order linear differential equation contains all possible solutions to the differential equation, usually with at least one arbitrary constant. This constant needs determination through additional information or conditions.
In this problem, the general solution is given as \(y(t) = Ce^{-10t} - 13\). This formula represents an entire family of solutions, determined by different values of \(C\).

• The general solution provides a comprehensive view of all potential functions that conform to the differential equation.
• It is a blueprint covering infinitely many curves or lines, each dictated by different constants like \(C\).

By substituting initial conditions into the general solution, we can distinguish the unique particular solution of interest. General solutions are indispensable in understanding the breadth of potential answers before narrowing down to a particular one through specific conditions.