Problem 2 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(1-10\) , use separation of variables to solve the initial value problem. Indicate the domain over which the solution is valid. \(\frac{d y}{d x}=-\frac{x}{y} \quad\) and \(y=3\) when \(x=4\)

Step-by-Step Solution

Answer

The solution of the initial value problem is given by \(y^{2} = -x^{2} + 10\), which is valid for \(-\sqrt{10} \leq x \leq \sqrt{10}\)

Step 1: Separate Variables

For the given differential equation, \(\frac{d y}{d x}=-\frac{x}{y}\) the variables can be separated by cross multiplying to give \(y \, dy = - x \, dx\)

Step 2: Integrate

Now, integrate both sides of the equation. This gives \(\int y \, dy = -\int x \, dx\), which can be integrated to yield \(\frac{y^{2}}{2} = -\frac{x^{2}}{2} + C\)

Step 3: Solve for C

Replace the initial values \(y=3\) when \(x=4\) into the equation: \(\frac{3^{2}}{2} = -\frac{4^{2}}{2} + C\). Solve for \(C\) and you find that \(C = 5\)

Step 4: Write out the general solution

Inserting \(C\) back into the equation yields the general solution: \(\frac{y^{2}}{2} = -\frac{x^{2}}{2} + 5\) or \(y^{2} = -x^{2} + 10\)

Step 5: Determine the domain

Since \(y\) is a real number, \(y^{2}\) should be a non-negative number, which implies that \(-x^{2} + 10 \geq 0\). Solve this inequality to find the domain: \(-x^{2} \geq -10\), yielding that \(x^{2} \leq 10\) or \(-\sqrt{10} \leq x \leq \sqrt{10}\)

Key Concepts

Separation of Variables

Separation of Variables is an essential technique used to solve ordinary differential equations. It works by isolating variables on opposite sides of an equation. In the exercise we are looking at, the given differential equation is \( \frac{dy}{dx} = -\frac{x}{y} \). Our goal is to separate \(x\) terms on one side and \(y\) terms on the other side.
This is done by multiplying both sides by \(y\) and \(dx\) giving us \(y \, dy = -x \, dx\).
This step is critical as it prepares the equation for integration.
Remember, ensuring variables are neatly separated is the foundation for solving these equations using this method.

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation with an additional condition called the initial condition. This tells us the value of the solution at a specific point. For example, in our problem, we know that \(y = 3\) when \(x = 4\). This condition is crucial as it helps in finding a specific solution from the family of solutions.
When the variables have been separated and integration is performed, the integration constant (often \(C\)) appears. This is where the initial condition comes into play.
We substitute the initial values given into the equation, allowing us to solve for \(C\).

• This initial condition uniquely determines the solution that fits the problem's specific scenario.

Integration

Integration is the mathematical process used to solve for accumulated values, such as areas under a curve. In differential equations, once variables have been separated, we use integration to find the antiderivatives.
For our separated equation \(y \, dy = -x \, dx\), integration gives us

• \(\int y \, dy = \frac{y^2}{2}\)
• \(-\int x \, dx = -\frac{x^2}{2}\)

Then we include an integration constant \(C\).
This process yields a general solution, which is a continuous expression that describes a family of potential solutions.
The initial condition then helps us find the particular solution by solving for \(C\).
Integration transforms our separated variables into a solvable equation.

Domain of Solutions

The domain of a solution is the set of input values for which the solution is defined. While dealing with real numbers, we are often interested in where the solution remains valid and meaningful.
In our equation \(y^2 = -x^2 + 10\), the domain is found by ensuring conditions where \(y^2\) is non-negative.
To find this, solve \(-x^2 + 10 \geq 0\), leading to

• \(-x^2 \geq -10\),
• thus \(x^2 \leq 10\),
• which results in \(-\sqrt{10} \leq x \leq \sqrt{10}\).

The domain tells us for which \(x\) values the solution \(y\) is real and valid. It is important to specify this range, as solutions outside this domain wouldn't be correct or applicable.
Determining the domain ensures clarity and accuracy in mathematical solutions.