Question

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

Short answer

Question: Solve the initial value problem for the given differential equation \(y^{\prime}(t)=\frac{e^{t}}{2 y}\) and initial condition \(y(\ln 2) = 1\). Answer: The solution for the initial value problem is \(y(t) = \sqrt{e^t - 1}\).

Step-by-step solution

Identify separable equation

We have the given differential equation: $$y^{\prime}(t)=\frac{e^{t}}{2 y}$$ Notice that this differential equation can be written as: $$y^{\prime}(t) = \frac{1}{2y} \cdot e^{t}$$ This shows that the equation is separable, as the left-hand side only depends on the dependent variable \(y\) and the right-hand side only depends on the independent variable \(t\).

Separate Variables

Now, separate the variables \(y\) and \(t\). To do this, we rewrite the equation as: $$2y y^{\prime}(t) = e^t$$ $$2y dy = e^t dt$$ Now, we have separated variables with \(y\)-terms on the left and \(t\)-terms on the right.

Integrate Both Sides

Next, integrate both sides of the equation: $$\int 2y dy = \int e^t dt$$ After integrating, we get: $$y^2 = e^t + C$$

Solve for \(y(t)\)

To find the general solution of the differential equation, solve for \(y\) in terms of \(t\): $$y(t) = \pm\sqrt{e^t + C}$$

Use Initial Condition

We are given the initial condition \(y(\ln 2) = 1\). Use this to solve for the constant \(C\): $$1 = \pm\sqrt{e^{\ln 2} + C}$$ $$1 = \pm\sqrt{2 + C}$$ For simplicity, choose the positive square root, which gives us: $$1 = \sqrt{2 + C}$$ Squaring both sides, we get: $$1 = 2 + C$$ Solve for \(C\): $$C = -1$$

Final Solution

Now, substitute the value of \(C\) back into the general solution to get the particular solution for the initial value problem: $$y(t) = \sqrt{e^t - 1}$$ And this is the solution of the initial value problem.

Key concepts

initial value problem

An initial value problem is a type of differential equation coupled with a specific value, known as the initial condition, for the dependent variable at a specified point. This helps identify a unique solution from an infinite number of possible solutions that a differential equation might have. When faced with an initial value problem, the process begins by solving the differential equation to find a general solution. Then, the initial condition is applied to find the value of any constants in the solution. For instance, in the given problem, the equation \(y^{\prime}(t)=\frac{e^{t}}{2 y}\) is paired with the initial condition \(y(\ln 2)=1\). After solving the differential equation, we use this condition to find the constant \(C\), which helps us achieve the specific solution \(y(t) = \sqrt{e^t - 1}\). This particular solution is tailored to match the initial condition at \(t = \ln 2\). Thus, the initial value problem provides direction, helping to pinpoint the particular solution that satisfies both the differential equation and the initial condition.

integration

Integration is a fundamental concept in calculus that involves finding a function given its derivative. It is often described as the opposite operation of differentiation. In the context of solving differential equations, integration is the step where we find functions from their rate of change. To solve the differential equation \(2y \ dy = e^t \ dt\), we integrate both sides separately. Each side of the equation is integrated with respect to its own variable. This process results in:

• \(\int 2y \, dy = y^2 + C_1\)
• \(\int e^t \, dt = e^t + C_2\)

Usually, we combine these constants, \(C_1\) and \(C_2\), into a single constant \(C\) when formulating solutions, like when we derived \(y^2 = e^t + C\) in our example.
By integrating, we move from the rate of change back to a functional expression of \(y\), allowing us to solve the equation and apply initial conditions for a specific solution.

solving differential equations

Solving differential equations involves finding a function that satisfies the given equation. Differential equations represent relationships between functions and their derivatives, and the solution is a function or a set of functions that fulfills this relationship. In many cases, as with our example, the solution involves more than just algebraic manipulation; it requires calculus techniques like integration and sometimes specific methods like separation of variables. By solving the equation \(2y \ dy = e^t \ dt\), we eventually uncover \(y^2 = e^t + C\). From here, solving for \(y\) involves algebraic techniques. We use the properties of equations and functions to isolate \(y\), leading to the potential solutions \(y = \pm\sqrt{e^t + C}\). Using initial conditions to simplify these to a single, particular solution is the final step, which turns generalized answers into specific and applicable solutions.

separation of variables

Separation of variables is a technique used to solve some differential equations, particularly those that are classified as 'separable'. A separable differential equation can be rewritten such that each variable and its derivatives are on different sides of the equation. The process begins by manipulating the given equation to isolate each variable. In the equation \(y^{\prime}(t) = \frac{1}{2y} \cdot e^{t}\), we rearrange to put terms involving \(y\) on one side and terms involving \(t\) on the other, resulting in \(2y \ dy = e^t \ dt\). This form allows us to integrate each side independently. Separation of variables provides a straightforward way to solve differential equations by breaking them down into simpler integral solutions for each variable. It's particularly useful for equations that naturally allow such separation with minimal rearrangement, simplifying the overall process of finding a solution. This technique is a cornerstone in calculus for solving differential equations, enabling us to tackle complex relations in simpler, more manageable parts.