Question

Solve the initial value problem. $$ y^{\prime}-2 y=2 y^{1 / 2}, \quad y(0)=1 $$

Short answer

Question: Solve the initial value problem (IVP) involving the first-order, non-linear ordinary differential equation: $$ y^{\prime}-2 y=2 y^{1 / 2}, \quad y(0)=1 $$ Answer: The solution to the initial value problem is given by the implicit equation: $$ \frac{1}{2}y + \ln|y^{1/2}| = t + \frac{1}{2} $$

Step-by-step solution

Separate variables

In order to separate variables, we need to have all \(y\) terms on one side and \(t\) terms on the other side. Rewrite the given ODE as: $$ y^{\prime} - 2y + 2y^{1/2} = 0 $$ Now, divide both sides by \(2y^{1/2}\): $$ \frac{y^{\prime}}{2y^{1/2}} - y^{-1/2} = 1 $$ Now we have separated the variables, with the \(y\) terms on the left side and the \(t\) terms on the right side.

Integrate both sides

Next, we need to integrate both sides of the equation with respect to \(t\). Let \(u = y^{1/2}\), then \(du = \frac{y^{\prime}}{2y^{1/2}}dt\), so: $$\int du - u^{-1}du = \int dt$$ Integrating both sides, we get: $$ \frac{1}{2}u^2 + \ln|u| = t + C_1 $$

Solve for y

Now, we need to solve for \(y(t)\). Since \(u = y^{1/2}\), we have: $$ \frac{1}{2}(y^{1/2})^2 + \ln|y^{1/2}| = t + C_1 $$ Simplifying, we obtain: $$ \frac{1}{2}y + \ln|y^{1/2}| = t + C_1 $$ This is an implicit equation for \(y(t)\). However, it is not easy to isolate \(y(t)\) in this equation, so we will use the initial condition \(y(0) = 1\) to proceed.

Use initial condition

Plug the initial condition \(y(0) = 1\) into the equation: $$ \frac{1}{2} \cdot 1 + \ln|1^{1/2}| = 0 + C_1 $$ Solving for \(C_1\), we find \(C_1 = \frac{1}{2}\): $$ \frac{1}{2}y + \ln|y^{1/2}| = t + \frac{1}{2} $$ While we cannot find an explicit expression for \(y(t)\) in this case, we have obtained an implicit solution to the initial value problem. The solution to the initial value problem is given by the equation: $$ \frac{1}{2}y + \ln|y^{1/2}| = t + \frac{1}{2} $$

Key concepts

Initial Value Problem

Initial value problems are essential in understanding how systems change over time. They consist of a differential equation together with a set condition, called the initial condition. This specific condition usually provides the value of the unknown function at a particular point. In our case, the differential equation is \( y' - 2y = 2y^{1/2} \) with the initial condition \( y(0) = 1 \).

• Differential Equation: Represents the relationship involving the derivatives of a function. In this exercise, it shows how the rate of change of \(y\) depends on \(y\) itself.
• Initial Condition: States that \(y(0) = 1\), providing a specific starting point that helps determine the exact behavior of \(y\) over time.

The initial condition is crucial because it allows us to solve for constants of integration that appear after integrating the differential equation.

Separation of Variables

Separation of variables is a powerful method used to solve first-order ordinary differential equations. The idea is to rearrange the equation so that each variable and its differentials appear on separate sides of the equation. For our differential equation \( y' - 2y = 2y^{1/2} \), we first rearrange it into the form \( y' - 2y + 2y^{1/2} = 0 \).
Next, by dividing through by \( 2y^{1/2} \), we get:\[\frac{y^{\prime}}{2y^{1/2}} - y^{-1/2} = 1\]Now the \( y \) terms are isolated on one side and \( t \) terms are on the other. This facilitates the process of integrating both sides separately, given the preparation for the integration step.

Implicit Solution

Implicit solutions arise when solving differential equations because not all solutions can be neatly expressed with the dependent variable isolated. In this problem, after performing integration and manipulation, we arrive at:\[\frac{1}{2}y + \ln|y^{1/2}| = t + C_1\]While it's often desirable to have an explicit solution, where the dependent variable (\(y\) in this case) is isolated, it's not always feasible to solve for \(y\) directly. Here, using the initial condition \( y(0) = 1 \) allows us to determine the constant \(C_1\).
Despite not obtaining \( y(t) \) as a standalone function, the implicit relation still perfectly describes the solution, given that \(y(t)\) satisfies this equation for all \(t\). Understanding implicit solutions can often be a gateway to numeric or graphical methods for further exploration.

Integration

Integration is the mathematical process inverse of differentiation and is crucial for solving differential equations. Once variables have been separated, the next step is to integrate both sides with respect to their variables. In our case, integration leads to:\[\int du - u^{-1}du = \int dt\]By integrating both sides, we derive the equation:\[\frac{1}{2}u^2 + \ln|u| = t + C_1\]Here, the right side simplifies directly to \(t + C_1\) because the integration of \( dt \) in terms of \( t \) is straightforward. We then substitute back the substituted variable \(u = y^{1/2}\) to relate back to \(y\).
Understanding integration in the context of differential equations helps solve for unknown constants of integration using initial conditions, thereby providing a specific solution to the problem.