Problem 48 - Advanced Engineering Mathematics

Question

Determine a plausible value of \(x_{0}\) for which the graph of the solution of the initial-value problem \(y^{\prime}+2 y=3 x-6, y\left(x_{0}\right)=0\) is tangent to the \(x\) -axis at \(\left(x_{0}, 0\right)\). Explain your reasoning.

Step-by-Step Solution

Answer

The plausible value of \(x_0\) is 1.

Step 1: Understanding the Problem

We need to determine a value for \(x_0\) where the given differential equation's solution is tangent to the \(x\)-axis at the point \((x_0, 0)\). This means the slope of the tangent, which is \(y'\), should be zero at that point.

Step 2: Solve the Differential Equation

The first step is to solve the differential equation \(y' + 2y = 3x-6\). Rewrite it in linear form and find an integrating factor, \(e^{\int 2 \, dx} = e^{2x}\). Multiply through the equation to get \(e^{2x} y' + 2 e^{2x} y = (3x-6) e^{2x}\). The left side is the derivative of \(e^{2x} y\), so we integrate both sides with respect to \(x\) to find \(e^{2x} y = \int (3x-6) e^{2x} \, dx\).

Step 3: Compute the Integral

To compute the integral \(\int (3x-6) e^{2x} \, dx\), use Integration by Parts, letting \(u = 3x-6\) and \(dv = e^{2x} \, dx\), then \(du = 3 \, dx\) and \(v = \frac{1}{2} e^{2x}\). This gives \(3x-6\) multiplied by \(\frac{1}{2} e^{2x}\); the integral becomes \(\frac{1}{2} (3x-6) e^{2x} - \frac{3}{2} \int e^{2x} \, dx\). Solve it to obtain \(e^{2x} y = \frac{3x-6}{2} e^{2x} - \frac{3}{4} e^{2x} + C\).

Step 4: Solve for \(y\)

Divide both sides by \(e^{2x}\) to solve for \(y\), yielding \(y = \frac{3x-6}{2} - \frac{3}{4} + Ce^{-2x}\).

Step 5: Apply Initial Condition

We apply the initial condition \(y(x_0)=0\). Substituting \(x = x_0\) into the equation, we get \(0 = \frac{3x_0 - 6}{2} - \frac{3}{4} + Ce^{-2x_0}\). This simplifies to find the constant \(C\), giving \(C = \frac{3}{4} - \frac{3x_0 - 6}{2}\) evaluated at \(x_0\).

Step 6: Condition for Tangency

The solution needs to be tangent to the \(x\)-axis at \((x_0, 0)\), so we require \(y'(x_0) = 0\). Differentiating \(y = \frac{3x-6}{2} - \frac{3}{4} + Ce^{-2x}\) gives \(y' = \frac{3}{2} - 2Ce^{-2x}\). Set \(y'(x_0)=0\) to find \(\frac{3}{2} - 2C e^{-2x_0} = 0\).

Step 7: Solve for \(x_0\)

Substitute \(C\) from the initial condition into the tangency condition: \(\frac{3}{2} = 2\left(\frac{3}{4} - \frac{3x_0 - 6}{2}\right) e^{-2x_0}\). Solve this equation for \(x_0\). After some algebra, you find this equation simplifies to \(x_0 = 1\).

Key Concepts

Initial-Value Problem

In differential equations, an "Initial-Value Problem" (IVP) consists of a differential equation together with a specified value, known as the initial condition. This initial condition is used at a starting point in order to determine a unique solution. In the given exercise, the initial-value problem is described by the equation:

• \( y' + 2y = 3x - 6 \)
• with the initial condition \( y(x_0) = 0 \) at \( (x_0, 0) \).

We are asked to find an \( x_0 \) such that the graph of the solution is tangent to the \( x \)-axis at this point. The initial-value problem helps in ensuring that our solution to the differential equation corresponds to a specific physical or geometrical situation.

Integration by Parts

"Integration by Parts" is a technique used in calculus to integrate products of functions. It is particularly useful when standard integration methods are insufficient. This technique stems from the product rule of differentiation and is expressed by:\[ \int u \, dv = uv - \int v \, du \]In solving the differential equation \( y' + 2y = 3x - 6 \), we encounter the integral \( \int (3x-6) e^{2x} \, dx \). By choosing \( u = 3x - 6 \) and \( dv = e^{2x} \, dx \), we calculate:

• \( du = 3 \, dx \)
• \( v = \frac{1}{2} e^{2x} \)

Applying integration by parts, we have:\[ \int (3x-6) e^{2x} \, dx = \frac{1}{2}(3x-6)e^{2x} - \frac{3}{2} \int e^{2x} \, dx \]The process simplifies the integral and aids in solving the differential equation.

Linear Differential Equation

A "Linear Differential Equation" is an equation that involves a function and its derivatives and is linear with respect to the unknown function and its derivatives. The form of our equation is:\[ y' + 2y = 3x - 6 \]The terms \( y' \) and \( y \) are both of the first degree, making this a linear equation. Solving such an equation often involves finding an integrating factor. In our case, the integrating factor is \( e^{\int 2 \, dx} = e^{2x} \).By multiplying the entire equation by this integrating factor, it transforms into a simpler form that allows us to find the solution by integrating both sides. This solving method is crucial as it converts a potentially complex problem into a manageable form.

Equation of Tangency

The "Equation of Tangency" refers to finding conditions under which a curve touches the line at a point without crossing it. For our differential equation solution to be tangent to the \( x \)-axis at \( (x_0, 0) \), the derivative \( y' \) must be zero at that point, as it implies a horizontal tangent.

• The derived form of the solution: \( y = \frac{3x-6}{2} - \frac{3}{4} + Ce^{-2x} \)
• To ensure tangency, set \( y'(x_0) = 0 \): \( y' = \frac{3}{2} - 2Ce^{-2x} \)

Applying these conditions, along with the given initial condition, lets us deduce that \( x_0 = 1 \) results in the solution being tangent at \( (x_0, 0) \). This assures that we have determined the correct point of tangency.