Question

Find the solution of the following initial value problems. $$y^{\prime}(t)=\frac{3}{t}+6 ; y(1)=8$$

Short answer

Question: Find the solution of the initial value problem given by the differential equation \(y^{\prime}(t) = \frac{3}{t} + 6\) with the initial condition \(y(1) = 8\). Answer: The solution of the given initial value problem is \(y(t) = 3 \ln |t| + 6t + 2\).

Step-by-step solution

Identify the differential equation

The given differential equation is $$ y^{\prime}(t)=\frac{3}{t}+6 .$$ This is the first-order and non-homogeneous differential equation.

Integrate the given equation

The equation is already in the separable form, so to find the general solution, directly integrate both sides of the equation with respect to \(t\): $$ \int y^{\prime}(t) \, dt = \int \left(\frac{3}{t} + 6\right) \, dt . $$

Evaluate the integral

Integrating the right side with respect to \(t\), we get: $$ y(t) = 3 \int \frac{1}{t} \, dt + 6 \int 1 \, dt .$$ Now, evaluating the integrals: $$ y(t) = 3 (\ln |t| + C_1) + 6 (t + C_2) .$$ Where \(C_1\) and \(C_2\) are constants of integration, combining them into a single constant, say \(C\): $$ y(t) = 3 \ln |t| + 6t + C .$$

Apply the initial condition

Apply the initial condition \(y(1) = 8\) to find the value of C: $$8 = 3 \ln |1| + 6(1) + C.$$ As \(\ln 1 = 0\), the equation becomes: $$8 = 6 + C .$$ So, the value of \(C\) is \(2\).

Write the particular solution

Substitute the value of \(C\) back into the general solution to get the particular solution of the initial value problem: $$ y(t) = 3 \ln |t| + 6t + 2 .$$ This is the solution of the given initial value problem.

Key concepts

Initial Value Problem

An initial value problem in the context of differential equations includes both a differential equation and an initial condition. The initial condition is a given value of the function that allows us to find a unique solution from the general family of solutions.
In this case, we have the differential equation \(y'(t) = \frac{3}{t} + 6\) and the initial condition \(y(1) = 8\). This condition is crucial as it helps pin down one specific solution by determining the constant of integration.
When you solve an initial value problem, your goal is to find the function \(y(t)\) that satisfies both the differential equation and the initial condition. If there was no initial condition, the solution would have an arbitrary constant appearing at the end. This constant is adjusted using the initial condition to meet that specific requirement of the solution passing through \((t, y) = (1, 8)\).

First-Order Differential Equations

First-order differential equations are equations that involve the first derivative of a function. Unlike higher-order equations, these equations do not have derivatives of higher orders like second (\(y''\)) or third (\(y'''\)) derivatives.
In the example given, \(y'(t) = \frac{3}{t} + 6\), the equation is first-order because it involves only the first derivative \(y'\). Determining the solution involves finding a function \(y(t)\) such that it, and its first derivative, satisfy this equation.
These types of differential equations are often easier to solve than higher-order ones. They require different techniques based on their form, such as separable equations or linear equations. In this exercise, the equation is separable, which simplifies the integration process as both sides can be integrated independently.

Integration Techniques

Integration techniques are essential for solving differential equations like the one in this exercise. To solve \(y'(t) = \frac{3}{t} + 6\), we integrate both sides with respect to \(t\) to find \(y(t)\).
Although this is a separable equation, general integration techniques apply here:

• Basic Integration: This involves integrating simple functions, such as \(\int \frac{1}{t} \, dt = \ln|t| + C\).
• Constant Integration: Integrating a constant like 6 gives \(\int 6 \, dt = 6t + C\).

The result is a combined solution \(y(t) = 3 \ln |t| + 6t + C\), which is adjusted using the initial condition. Understanding these integration techniques is key to solving differential equations effectively.

Separable Differential Equations

Separable differential equations are a specific type of differential equation where the variables and their derivatives can be separated on opposite sides of the equation. This separation simplifies the process, allowing each side to be integrated independently.
In this exercise, rearranging \(y'(t) = \frac{3}{t} + 6\) allows us to integrate directly after recognizing that each component involving \(t\) is already on one side.

• Separating involves writing the equation in the form \( \frac{dy}{dt} = f(t)g(y)\), and then manipulating it into \( g(y)dy = f(t)dt \).
• This approach culminates in simultaneously integrating both sides to find a solution that includes a constant of integration, later resolved by the initial condition.

The separability allows for flexibility and simplification in solving various first-order equations, showcasing this method's power and efficiency.