Question

Find a good approximation, for \(x\) large and positive, to the solution of the equation $$ y^{\prime \prime}-\frac{3}{x} y^{\prime}+\left(\frac{15}{4 x^{2}}+x^{1 / 2}\right) y=0 $$ Hint: Remove first derivative term.

Short answer

For large \( x \), the solution is approximately \( y(x) \approx C_1 \text{cos}(x^{1/4}) + C_2 \text{sin}(x^{1/4}) \).

Step-by-step solution

- Identify dominant terms

For large and positive values of \( x \), identify dominant terms in the given differential equation. The dominant term in the equation is likely going to be the one with the highest power of \( x \).

- Simplify the equation

Given the hint, simplify the differential equation by ignoring the term with the first derivative, \( -\frac{3}{x} y' \). The simplified equation becomes: \[ y'' + \left( \frac{15}{4 x^2} + x^{1/2} \right) y = 0 \]

- Approximations for large \( x \)

For large \( x \), the term \( x^{1/2} \) becomes significantly larger compared to \( \frac{15}{4 x^2} \). Hence, approximate the equation as: \[ y'' + x^{1/2} y = 0 \]

- Find a trial solution

Try a solution of the form \( y = e^{kx} \). Compute the first and second derivatives: \( y' = ke^{kx} \) and \( y'' = k^2 e^{kx} \). Substitute these into the approximated equation.

- Substitute and solve for \( k \)

Substitute \( y'' = k^2 e^{kx} \) and \( y = e^{kx} \) into the approximated equation: \[ k^2 e^{kx} + x^{1/2} e^{kx} = 0 \] Divide by \( e^{kx} \): \[ k^2 + x^{1/2} = 0 \] Solve for \( k \): \[ k^2 = -x^{1/2} \] Since \( k \) must be imaginary, let \( k = i \beta \), so \( \beta^2 = x^{1/2} \) and \( \beta = \frac{x^{1/4}}{\text{i}} \).

- General solution

The general solution to the differential equation is: \[ y(x) = C_1 e^{i x^{1/4}} + C_2 e^{-i x^{1/4}} \] Using Euler's formula, this can be written as: \[ y(x) = C_1 \text{cos}(x^{1/4}) + C_2 \text{sin}(x^{1/4}) \]

Key concepts

Second-order differential equations

Second-order differential equations involve second derivatives, i.e., terms like y''. In our equation, we are dealing with a second-order differential equation in the form: $$y^{\text{''}} - \frac{3}{x}y^\text{'} + \frac{15}{4x^2}y + x^{1/2}y = 0$$We are given the hint to remove the first derivative term, making our task slightly easier. Second-order differential equations often describe physical phenomena such as oscillations and heat conduction, making them essential in various scientific fields.

Asymptotic analysis

Asymptotic analysis is useful for approximating solutions to differential equations, especially for large values of the variable. In this problem, we're interested in how our solution behaves when x is large and positive.### Dominant TermsFor large x values, we focus on the dominant terms in the equation. The term with the highest power of x or the fastest growth rate generally dominates.### SimplificationIgnoring the less significant terms simplifies our calculations. For our equation, we can ignore the first derivative term as suggested, leading to:$$y^{''} + \frac{15}{4x^{2}}y + x^{1/2}y = 0$$Simplified further for large x, the equation approximates to:$$y^{''} + x^{1/2}y = 0$$

Large x approximation

Large x approximations allow us to simplify complex equations by focusing on the most significant terms.### Transition to ApproximationGiven that for very large x, the term x^{1/2} will be notably larger than \frac{15}{4x^{2}}, we approximate:$$y^{''} + x^{1/2}y = 0$$### Trial SolutionWe can use a trial solution of the form y = e^{kx}. By substituting y = e^{kx} and its derivatives (y' = ke^{kx}, y'' = k^2e^{kx}) into our approximated equation, we get:$$k^{2}e^{kx} + x^{1/2}e^{kx} = 0$$### Solving for kDivide by e^{kx} and solve for k, leading to:$$k^{2} + x^{1/2} = 0$$Given k must be imaginary, set k = i \beta. Consequently, \beta^{2} = x^{1/2}, so \beta = \frac{x^{1/4}}{\text{i}}.

Euler's formula

Euler's formula is pivotal in connecting exponential functions with trigonometric functions. It's stated as:$$e^{ix} = \text{cos}(x) + i\text{sin}(x)$$### Applying Euler's FormulaIn our solution, we have imaginary components which Euler's formula helps simplify.Thus, the general solution for our differential equation transforms from:$$y(x) = C_1e^{ix^{1/4}} + C_2e^{-ix^{1/4}}$$to a more familiar form using Euler's formula:$$y(x) = C_{1}\text{cos}(x^{1/4}) + C_{2}\text{sin}(x^{1/4})$$This representation is easier to interpret and apply in many physical contexts.