Question

Consider the initial value problem $$ y^{\prime}+\frac{2}{3} y=1-\frac{1}{2} t, \quad y(0)=y_{0} \text { . } $$ Find the value of \(y_{0}\) for which the solution touches, but does not cross, the \(t\) -axis.

Short answer

The value of y0 for which the solution touches but does not cross the t-axis is \(y0 = \frac{3}{2}+\frac{9}{8}\).

Step-by-step solution

Identify the ODE as linear and first-order

The given ODE is of the form $$ y^\prime + P(t)y = Q(t), $$ where \(P(t) = \frac{2}{3}\) and \(Q(t) = 1 - \frac{1}{2}t\). This is a linear first-order ODE.

Find the integrating factor

To find the integrating factor, we first need to find the function \(\mu(t)\). The \(\mu\) function is given by $$ \mu(t) = e^{\int P(t)dt}. $$ Here, \(P(t) = \frac{2}{3}\), so $$ \mu(t) = e^{\int \frac{2}{3} dt} = e^{\frac{2}{3}t}. $$

Multiply the ODE by the integrating factor

Multiply both sides of the given ODE by the integrating factor \(\mu(t) = e^{\frac{2}{3}t}\): $$ e^{\frac{2}{3}t}y^\prime + \frac{2}{3}e^{\frac{2}{3}t}y = \left(1 - \frac{1}{2}t\right)e^{\frac{2}{3}t}. $$

Integrate both sides with respect to t

Now, integrate both sides with respect to t: $$ \int \left(e^{\frac{2}{3}t}y^\prime + \frac{2}{3}e^{\frac{2}{3}t}y\right) dt = \int \left(1 - \frac{1}{2}t\right)e^{\frac{2}{3}t} dt. $$ The left side of the equation is the derivative of \(e^{\frac{2}{3}t}y\) and therefore integrating it gives: \(e^{\frac{2}{3}t}y\). To integrate the right side, we can use integration by parts: Let \(u = (1 - \frac{1}{2} t)\) and \(dv = e^{\frac{2}{3}t}dt\). Then, we have \(du = -\frac{1}{2}dt\) and \(v = \frac{3}{2}e^{\frac{2}{3}t}\). Using integration by parts, we get: $$ \int \left(1 - \frac{1}{2}t\right)e^{\frac{2}{3}t} dt = \left(1 - \frac{1}{2}t\right)\frac{3}{2}e^{\frac{2}{3}t} - \int \frac{3}{2}e^{\frac{2}{3}t}\left(-\frac{1}{2}\right)dt, $$ which simplifies to: $$ \int \left(1 - \frac{1}{2}t\right)e^{\frac{2}{3}t} dt = \left(1 - \frac{1}{2}t\right)\frac{3}{2}e^{\frac{2}{3}t} + \frac{3}{4}\int e^{\frac{2}{3}t}dt = \left(1 - \frac{1}{2}t\right)\frac{3}{2}e^{\frac{2}{3}t} + \frac{9}{8}e^{\frac{2}{3}t}. $$ From these integrations, we get: $$ e^{\frac{2}{3}t}y = \left(1 - \frac{1}{2}t\right)\frac{3}{2}e^{\frac{2}{3}t} + \frac{9}{8}e^{\frac{2}{3}t} + C, $$ where C is the constant of integration.

Solve for y(t)

Divide both sides by \(e^{\frac{2}{3}t}\) to find the general solution for y(t): $$ y(t) = \left(1 - \frac{1}{2}t\right)\frac{3}{2} + \frac{9}{8} + Ce^{-\frac{2}{3}t}. $$

Apply the initial condition y(0) = y0

To find the value of C, plug in the initial condition, \(y(0) = y0\): $$ y0 = \left(1 - \frac{1}{2}(0)\right)\frac{3}{2} + \frac{9}{8} + Ce^{-\frac{2}{3}(0)}. $$ Simplifying, we get: $$ y0 = \frac{3}{2} + \frac{9}{8} + C. $$ Solving for C, we have: $$ C = y0 - \frac{3}{2} - \frac{9}{8}. $$ Putting the value of C in the general solution, we get: $$ y(t) = \left(1 - \frac{1}{2}t\right)\frac{3}{2} + \frac{9}{8} + \left(y0 - \frac{3}{2} - \frac{9}{8}\right)e^{-\frac{2}{3}t}. $$

Find y0 such that y(t) touches but does not cross the t-axis

We know that y(t) touches the t-axis when \(y(t) = 0\). Let's solve for y0 by setting y(t) equal to zero and t equal to a specific value \(t_1\): $$ 0 = \left(1 - \frac{1}{2}t_1\right)\frac{3}{2} + \frac{9}{8} + \left(y0 - \frac{3}{2} - \frac{9}{8}\right)e^{-\frac{2}{3}t_1}. $$ We can see that the function inside the exponent, \(e^{-\frac{2}{3}t_1}\), is always positive. This means that the solution will neither touch nor cross the t-axis if the term inside the parentheses is not equal to zero. Hence: $$ y0 - \frac{3}{2} - \frac{9}{8} = 0. $$ Solving for y0, we get $$ y0 = \frac{3}{2} + \frac{9}{8}. $$ The value of y0 for which the solution touches but does not cross the t-axis is \(y0 = \frac{3}{2}+\frac{9}{8}\).

Key concepts

Linear First-Order ODE

A linear first-order ordinary differential equation (ODE) is a pivotal concept in calculus and its applications. It represents a relationship where the rate of change of a function, denoted as y'(t), is a linear function of the variable y itself.

An important characteristic of these ODEs is their standard form, which is written as:
\[ y'(t) + P(t)y = Q(t) \]
Here, P(t) and Q(t) are known functions of the independent variable t. The solution to a linear first-order ODE involves finding the function y(t) that satisfies this equation under given conditions.

Our textbook problem falls into this category, with P(t) equating to \(\frac{2}{3}\) and Q(t) to \(1 - \frac{1}{2}t\). Understanding the structure of the ODE is crucial because it guides us to the appropriate method for finding its solution, which, in this case, involves the use of an integrating factor.

Integrating Factor

The integrating factor is an ingenious technique used to solve linear first-order ODEs. It involves creating a new function μ(t) that transforms the original ODE into an exact differential equation, making it more manageable to integrate and solve.

The formula for the integrating factor is:
\[ \mu(t) = e^{\int P(t)dt} \]
Applying this to our problem, with P(t) being a constant \(\frac{2}{3}\), we find the integrating factor to be \( e^{\frac{2}{3}t} \). By multiplying the entire ODE by this factor, we facilitate the integration process, because it turns the left side of the equation into the derivative of \(e^{\frac{2}{3}t} y\), which can be integrated directly.

It's the integrating factor that allows us to proceed from the given ODE to its solution, by cleverly simplifying the equation. The proper choice of integrating factor is critical as it ensures the solvability of an ODE where direct integration is not feasible.

Boundary Conditions

Boundary conditions in differential equations are additional pieces of information that specify the value of the solution at a certain point. They are essential for determining the specific solution of an ODE from a family of possible solutions.

In initial value problems like the one given, the boundary condition is provided in the form of an initial value, y(0) in our case, labeled as y0. This initial condition anchors the solution to a specific starting point. Utilizing the initial condition allows us to solve for the constant of integration C that arises when integrating the ODE.

For the solution to touch but not cross the t-axis, we exploit the fact that boundary conditions can tell us where the function y(t) achieves certain behaviors — in this problem, where y(t) equals zero. From this, we deduce the value of y0 such that the solution graph just grazes the t-axis without dipping below. This integration constant and boundary condition interplay is crucial for modeling real-world phenomena where initial states often dictate the evolution of a system.