Question

Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem. $$t y^{\prime}(t)=1, y(1)=2$$

Short answer

Question: Solve the given Initial Value Problem (IVP) with a first-order differential equation and determine the solution: $$t\frac{dy}{dt} = 1$$ Initial condition: \(y(1) = 2\). Answer: The solution to the given initial value problem is: $$y(t) = \ln|t| + 2$$

Step-by-step solution

Check if the equation is separable

First, let's rewrite the given equation in a clear form, remembering that \(y^{\prime}(t)=\frac{dy}{dt}\): $$t\frac{dy}{dt} = 1$$ We can rewrite the equation (once rearranged) in the form \(\frac{dy}{y} = g(t)dt\), which will indicate that the equation is separable.

Rearrange the equation for separation of variables

We want to separate the variables \(y\) and \(t\), so we need to rewrite the equation as a product of a function of \(y\) and a function of \(t\). We achieve this by dividing both sides by \(t\): $$\frac{dy}{dt} = \frac{1}{t}$$ Now we can separate the variables by multiplying both sides by \(dt\), and we obtain the following equation: $$dy = \frac{1}{t}dt$$

Integrate both sides of the separable equation

We'll now integrate both sides of the separable equation with respect to their respective variables: $$\int dy = \int \frac{1}{t}dt$$ The integral of \(dy = y + C_1\) and the integral of \(\frac{1}{t}dt = \ln|t| + C_2\) (where \(C_1\) and \(C_2\) are constants of integration).

Combine the constants of integration

We have the following after integrating: $$y + C_1 = \ln|t| + C_2$$ To simplify the equation, we'll combine the constants of integration into one single constant, \(C = C_1 - C_2\). Thus, our equation becomes: $$y = \ln|t| + C$$

Apply initial conditions

We are given the initial condition that \(y(1) = 2\). We'll substitute these values into the equation to find the value of the constant \(C\): $$2 = \ln|1| + C$$ Since \(\ln|1| = 0\), we have \(C = 2\).

Write the solution of the initial value problem

Now we have the complete solution for the IVP: $$y(t) = \ln|t| + 2$$

Key concepts

Separable Differential Equations

Separable Differential Equations involve assigning differential equations into a form where the variables can be separated. This concept is key when the goal is to solve an initial value problem, as shown in the given exercise. The original problem, \( t\frac{dy}{dt} = 1 \), must be transformed into a format that allows variables of \( y \) and \( t \) to be separated on opposite sides of the equation. This is done by rearranging to the form \( \frac{dy}{y} = g(t)dt \). This separation helps us handle the integration process later on more easily.

Symbolically, if a function can be written as \( \frac{dy}{dx} = f(y)g(x) \), it is deemed to be separable. Breaking down each part of the equation helps when integrating each side with respect to their respective variables. After separation, we can solve each side individually, preparing us to find a more workable solution to the differential equation.

Separation of Variables

Separation of Variables is a powerful technique for solving differential equations. Once you've confirmed that an equation is separable, the next step is implementing separation of the variables. This involves rewriting the equation so that each variable is on one side. In the current exercise, this was achieved by dividing both sides by \( t \), getting to \( \frac{dy}{dt} = \frac{1}{t} \).

After simplification, multiplying both sides by \( dt \) results in \( dy = \frac{1}{t}dt \), effectively separating \( y \) and \( t \).
By achieving this setup, the integrals of these isolated expressions can now be evaluated. This step breaks the problem down into smaller, more manageable components. This is crucial for making further analysis and equation solving feasible. Essentially, this method simplifies complicated interrelationships in a differential equation making them easier and more straightforward to solve when aiming for a function of \( y \) that satisfies the given conditions.

Integration Methods

Integration Methods are employed to evaluate the integrals once the variables in a differential equation have been successfully separated. In our example, we have two integrals to consider: \( \int dy \) and \( \int \frac{1}{t} dt \).

Selecting appropriate integration techniques based on the form of functions multiplying the differentials is essential.

• For \( \int dy \), the integration yields \( y + C_1 \), where \( C_1 \) is a constant of integration.
• For \( \int \frac{1}{t} dt \), the result is \( \ln|t| + C_2 \).

Combining these results into a single equation with a condensed constant \( C = C_1 - C_2 \), simplifies our expression to \( y = \ln|t| + C \). Applying initial conditions, such as \( y(1) = 2 \), lets us solve for \( C \), establishing the unique solution \( y(t) = \ln|t| + 2 \). These are the nuts and bolts of solving an initial value problem with a separable differential equation.