Question

(a) Obtain an implicit solution and, if possible, an explicit solution of the initial value problem. (b) If you can find an explicit solution of the problem, determine the \(t\)-interval of existence. $$ y \frac{d y}{d t}-\sin t=0, \quad y(\pi / 2)=-2 $$

Short answer

Question: Find the implicit and explicit solution to the initial value problem \(y \frac{d y}{d t}-\sin t=0\) with initial condition \(y(\pi/2)=-2\) and determine the \(t\)-interval of existence. Answer: The implicit solution is \(\ln(|y|)=-\cos t + \ln2\). The explicit solution is \(y=-2e^{-\cos t}\). The \(t\)-interval of existence is \((-\infty, \infty)\).

Step-by-step solution

Rewrite the given equation in the right form

The given differential equation is $$y \frac{d y}{d t}-\sin t=0.$$ To rewrite it in a suitable form, let's add \(\sin t\) to both sides of the equation to get: $$y\frac{dy}{dt}=\sin t.$$

Solve the first-order differential equation using separation of variables

To solve the differential equation, we will separate the variables, moving all \(y\) terms to one side and all \(t\) terms to the other side: $$\frac{dy}{y}=\frac{\sin t}{1}dt.$$ Now, integrate both sides with respect to their respective variables: $$\int\frac{dy}{y}=\int\sin t dt.$$

Apply the initial condition to determine the constant of integration

After integrating both sides, we obtain: $$\ln(|y|)=-\cos t+C,$$ where \(C\) is the constant of integration. Applying the initial condition \(y(\pi/2)=-2\), we can solve for \(C\): $$\ln|-2|=-\cos(\pi/2)+C$$ $$\ln2=C$$.

Find an explicit solution if possible

Now that we have the constant of integration, substitute it back into the equation and solve for \(y\): $$\ln(|y|)=-\cos t + \ln2$$ Exponentiate both sides: $$|y|=e^{-\cos t+\ln2}$$ $$y = \pm 2e^{-\cos t}$$ Using the initial condition, \(y(\pi/2)=-2\), we can find that the correct explicit solution is: $$y=-2e^{-\cos t}$$.

Determine the t-interval of existence

Now that we have an explicit solution, we need to determine the t-interval of existence. Since the exponential function is continuous for all real numbers and the cosine function is also continuous for all real numbers, there are no restrictions on the existence of the solution. Therefore, the \(t\)-interval of existence is \((-\infty, \infty)\).

Key concepts

Implicit Solution

An implicit solution of a differential equation defines a relation between the dependent variable and the independent variable without explicitly providing a formula for the dependent variable in terms of the independent one. In other words, you have an equation involving both variables that holds true, rather than a direct equation like \(y = f(t)\). It is important in scenarios where solving explicitly is difficult or impossible. In our original problem, the implicit solution was formed after the integration step: \[\ln(|y|) = -\cos(t) + C\]This relation shows the connection between \(y\) and \(t\), although \(y\) is not isolated on one side. Implicit solutions are

• Useful when explicit forms are too complicated or non-existent
• Best utilized in the initial phases of solving differential equations

Understanding implicit solutions is essential when you begin the process of solving initial value problems.

Explicit Solution

An explicit solution transforms the implicit relation into a straightforward expression where the dependent variable is isolated in terms of the independent variable. This makes calculations and interpretations much easier. From our exercise, after using the initial condition, we found the explicit solution:\[ y = -2e^{-\cos(t)} \]Here, \(y\) is clearly defined as a function of \(t\). This explicit form is often desired as it enables us to compute values directly.Benefits of explicit solutions include:

• Easier to use for predictions and calculations
• Facilitates understanding and visualization of the relationship

Achieving an explicit solution can sometimes require additional steps like using initial conditions or extra computations.

Interval of Existence

The interval of existence refers to the range of the independent variable (typically \(t\)) over which a solution to a differential equation is valid. This concept ensures that no part of the solution extends beyond the boundaries defined by mathematical constraints such as division by zero or negative roots where undefined.In our solved exercise, we determined the \(t\) interval of existence to be \((-\infty, \infty)\). This conclusion arises because both the exponential and cosine functions involved in the solution are continuous for all real numbers.To find the interval of existence, consider:

• Points where functions might be undefined (like division by zero)
• Any discontinuities within the function set

Understanding the interval helps in applying the solution correctly within its limits.

Separation of Variables

Separation of variables is a fundamental method for solving first-order differential equations. The idea is to manipulate the equation such that all terms involving the dependent variable \(y\) and its derivative are on one side, and all terms involving the independent variable \(t\) are on the other side.This method was employed in our original solution:\[\frac{dy}{y} = \frac{\sin t}{1} dt\]Steps for using this method effectively:

• Rearrange the equation to isolate derivatives and assign constants appropriately.
• Integrate both sides with respect to their variables.
• Apply any given initial conditions to resolve constants of integration.

Separation of variables simplifies solving complex equations step-by-step and leads you towards finding solutions, implicit or explicit.