Problem 25 - Advanced Engineering Mathematics

Question

Solve the given initial-value problem. Use a graphing utility to graph the solution curve. $$ x^{2} y^{\prime \prime}+3 x y^{\prime}=0, y(1)=0, y^{\prime}(1)=4 $$

Step-by-Step Solution

Answer

The solution is \( y(x) = -\frac{2}{x^{2}} + 2 \).

Step 1: Analyze the differential equation

The given differential equation is \( x^{2} y'' + 3xy' = 0 \). This is a second-order linear homogeneous differential equation with constant coefficients. Our goal is to solve this equation using appropriate methods.

Step 2: Transform into a simpler form

Let \( y' = v \), which implies that \( y'' = v' \). Substitute these into the differential equation: \( x^{2} v' + 3xv = 0 \). This is a first-order linear differential equation in terms of \( v(x) \).

Step 3: Solve the first-order differential equation

Rearrange the equation as \( v' + \frac{3}{x}v = 0 \). This is a separable equation. Separate variables: \( \frac{dv}{v} = -\frac{3}{x}dx \). Integrate both sides: \( \ln|v| = -3 \ln|x| + C \). Exponentiate to isolate \( v \): \( v = \frac{C}{x^{3}} \).

Step 4: Solve for \( y \) using integration

Recall \( v = y' = \frac{C}{x^{3}} \). Integrate to find \( y \): \( y = \int \frac{C}{x^{3}} \, dx = -\frac{C}{2x^{2}} + D \).

Step 5: Apply initial conditions

Use the initial conditions \( y(1) = 0 \) and \( y'(1) = 4 \):1. From \( y(1) = 0 \), substitute into \( y = -\frac{C}{2x^{2}} + D \): \( 0 = -\frac{C}{2 \cdot 1^{2}} + D \) results in \( D = \frac{C}{2} \).2. From \( y'(1) = 4 \), substitute into \( y' = \frac{C}{x^{3}} \): \( 4 = \frac{C}{1^{3}} \), which means \( C = 4 \).Thus, \( D = \frac{4}{2} = 2 \).

Step 6: Write the final solution

Substitute \( C \) and \( D \) back into the expression for \( y \):\[ y(x) = -\frac{4}{2x^{2}} + 2 = -\frac{2}{x^{2}} + 2. \]

Step 7: Graph the solution curve

Using a graphing utility, plot the solution \( y(x) = -\frac{2}{x^{2}} + 2 \). Observe that it represents a curve that approaches \( y = 2 \) as \( x \to \infty \) and has a vertical asymptote at \( x = 0 \).

Key Concepts

Linear Differential Equations

Linear differential equations are a type of differential equation where the dependent variable and its derivatives appear linearly. They can either be ordinary or partial, depending on whether they involve functions of only one variable or multiple variables, respectively. In simpler terms, a linear differential equation might sound complex, but essentially, it means the function and its derivatives are only multiplied by constants or functions of the independent variable.
In our exercise, the equation \( x^2 y'' + 3xy' = 0 \) is a second-order linear differential equation. To solve it, we need to use techniques like substitution and separation of variables, which simplify the system and make it easier to handle.
These equations are incredibly important in physics and engineering, where they describe systems from electromagnetic theory to thermal dynamics. Understanding how to approach and solve them is a crucial skill for students studying these fields.

Homogeneous Differential Equations

A homogeneous differential equation is a type where all terms are of the same degree. Specifically, for a linear homogeneous differential equation, all terms are a multiple of the function or its derivatives.
In simple words, if you have derivatives and constants like our example, and when you set the equation to zero, you have a homogeneous equation. In the equation \( x^2 y'' + 3xy' = 0 \), it qualifies as homogeneous because both terms involve derivatives of \( y \), and there are no standalone constant or function terms.
One thing to remember is when dealing with homogeneous equations, the solutions typically reflect a particular structure, which can be beautifully exploited using methods like separation of variables or substitution, as done in the provided solution steps.

Initial Conditions

Initial conditions are specific values provided in an initial-value problem that are used to find particular solutions. For differential equations, they typically specify the value of the function and its derivatives at a particular point.
In our scenario, the initial conditions were given as \( y(1) = 0 \) and \( y'(1) = 4 \), which means that at \( x = 1 \), the value of the function \( y \) is 0 and its derivative is 4. These conditions are crucial because they allow us to find the specific constants in our general solution, transforming it into a unique solution that fits the problem exactly.
Solving the problem without applying these conditions would only give us a general solution, which might explain the behavior of a family of solutions rather than pinpointing the exact curve that fits our specific initial conditions.

Separable Equations

Separable equations are a class of differential equations that can be separated into two integrals, each involving a different variable. This separation allows us to integrate and solve them more easily by breaking them into simpler parts.
In the exercise, the step of rewriting the equation \( v' + \frac{3}{x}v = 0 \) into a separable form \( \frac{dv}{v} = -\frac{3}{x}dx \) allowed us to integrate both sides individually. This technique is one of the first steps students learn because of its straightforward and effective approach in handling certain ordinary differential equations.
By transforming a complex problem into two simpler parts, we can integrate and find solutions that are otherwise more cumbersome to solve directly. Separable equations simplify the job by letting each variable live in its world during the integration step.