Question

Some of the results requested in Problems 21 through 28 can be obtained either by solving the given equations analytically, or by plotting numerically generated approximations to the solutions. Try to form an opinion as to the advantages and disadvantages of each approach. Solve the initial value problem $$ y^{\prime}=\left(1+3 x^{2}\right) /\left(3 y^{2}-6 y\right), \quad y(0)=1 $$ and determine the interval in which the solution is valid. Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent.

Short answer

Question: Given the first-order ordinary differential equation (ODE) initial value problem: \(\frac{dy}{dx} = \frac{3y^2 - 6y}{y^2}\) and its initial condition: \(y(0) = 1\), find the solution and its interval of validity. Answer: The solution to the ODE is given by \(3y - 6\ln|y| = x + x^3 + 3\). The interval of validity is \(0 < y < 2\).

Step-by-step solution

Separate the variables and integrate

The given ODE is a first-order, separable ODE since we can re-write it in the form: \[ \left(\frac{3y^2 - 6y}{y^2}\right)dy = (1 + 3x^2)dx. \] Now we can integrate both sides: \[ \int\left(\frac{3y^2 - 6y}{y^2}\right)dy = \int(1 + 3x^2)dx. \] Integrating both sides, we get: \[ \Rightarrow \int (3 - \frac{6}{y}) dy = \int (1 + 3x^2) dx, \] \[ \Rightarrow 3y - 6\ln|y|= x + x^3 + C. \]

Apply the initial condition

We are given the initial condition \(y(0) = 1\). Applying this to our expression, we can solve for the constant \(C\): \[ 3(1) - 6\ln|1| = 0 + 0 + C, \] \[ \Rightarrow C = 3. \] Thus, our solution is: \[ 3y - 6\ln|y| = x + x^3 + 3. \]

Determine the interval of validity

To find the interval of definition, we look for points where the integral curve has a vertical tangent. This occurs when the denominator of the right-hand side of the original ODE equals zero: \[ 3y^2 - 6y = 0 \Rightarrow 3y(y - 2) = 0. \] The vertical tangent occurs for \(y = 0\) and \(y = 2\). Since \(y(0) = 1\), this lies between \(y = 0\) and \(y = 2\). Therefore, the interval of validity is: \[ 0 < y < 2. \] By taking into account the initial condition \(y(0) = 1\), the solution is valid as long as \(0 < y < 2\).

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) in differential equations consists of finding a solution to a differential equation that satisfies a given initial condition. In simpler terms, it means solving the differential equation and ensuring the solution meets a specific starting point or value. Our exercise begins with the differential equation: \( y^{\prime} = \frac{1+3x^{2}}{3y^{2}-6y} \) and the initial condition is provided as \( y(0) = 1 \).

• The IVP asks us to find a function \( y(x) \) that is not only a solution of the differential equation, but also satisfies \( y(0) = 1 \).
• To solve it, we first separate variables and integrate, allowing us to apply the initial condition to solve for any constants.

Applying the initial condition helps us pin down this constant and ensures that out of potentially many solutions, we have the one that fits our starting point in the exercise.

Interval of Validity

The Interval of Validity defines the range of \( x \) over which the solution to a differential equation is valid. In our exercise, we use the concept of vertical tangents to find where the solution is valid by evaluating when and where the equation itself loses definition.

• In the given equation \( 3y^{2} - 6y = 0 \), setting this denominator to zero shows where vertical tangents might occur.
• By solving \( 3y(y-2) = 0 \), we find potential critical points at \( y = 0 \) and \( y = 2 \).

Using the initial condition \( y(0) = 1 \), we ensure that the initial solution is located within the bounds \( y = 0 \) and \( y = 2 \). Thus, the interval of validity exists where the solution consistently holds true, specifically between these boundaries. This gives us the interval \( 0 < y < 2 \), where the solution doesn't crash into undefined values, retaining its mathematical correctness.

Vertical Tangent

A Vertical Tangent appears in a differential equation solution when the slope of the curve becomes infinite, typically at points where a denominator in the equation equals zero. Vertical tangents in differential equations are crucial to understanding where a solution might not be valid or could change behavior drastically.

• The notion is key when determining the interval of validity by preventing undefined behavior in solutions.
• In our exercise, we examined where \( 3y^{2} - 6y = 0 \) provides such tangential behavior, isolating values of \( y \) that lead to vertical tangents.

Vertical tangents can appear at any junction where the differential equation loses its regular differential form due to zero denominator values. This phenomenon helps math enthusiasts narrow down ranges for correct solutions, ensuring mathematical validity by dodging potentially infinite slopes or undefined points.