Question

\(x y^{\prime \prime}+5 y^{\prime}+x y=0\)

Short answer

The general solution is \( y(x) = C_1 + \frac{C_2}{x^4} \).

Step-by-step solution

- Identify the form of the differential equation

The given differential equation is \[x y^{\prime \prime}+5 y^{\prime}+x y=0\]Identify that it is a second-order linear homogeneous differential equation with variable coefficients.

- Look for a suitable method of solving

Since the coefficients are all functions of x, attempt solving this using a possible series solution or assuming solutions of the form based on characteristic equations if there are specific substitutions.

- Substitution and simplification

Assume a solution of the form \[y(x) = x^r\]Substitute this into the differential equation to determine the characteristic equation.

- Find the first and second derivatives

Calculate \[y^{\prime} = r x^{r-1}\]and\[y^{\prime\prime} = r(r-1) x^{r-2}\].

- Substitute derivatives back into the equation

Substitute \[y = x^r, y^{\prime} = r x^{r-1}\]and \[y^{\prime\prime} = r(r-1) x^{r-2}\]into \[x y^{\prime\prime}+5 y^{\prime}+x y=0\]which yields \[x (r(r-1) x^{r-2}) + 5 (r x^{r-1}) + x (x^r) = 0\].

- Simplify the equation

Simplify to get \[r(r-1) x^r + 5 r x^r + x^{r+1} = 0\].

- Factor out common term

Factor out \(x^r\) to get \[x^r (r(r-1) + 5r + x) = 0\].

- Set characteristic equation to zero

Solve for the characteristic equation: \[r(r-1) + 5r + x = 0\].

- Solve the characteristic equation

Note that for non-trivial solutions, \[x = 0\] implies x goes to zero or r satisfies \[r^2 + 4r = 0\]. Solve to get the roots: \[r(r+4) = 0\], hence \[r = 0\] or \[r = -4\].

- Form the general solution

The general solution from the roots of the characteristic equation is \[y(x) = C_1 x^0 + C_2 x^{-4} = C_1 + \frac{C_2}{x^4}\].

Key concepts

Homogeneous Differential Equation

A homogeneous differential equation is one where every term is a function of the dependent variable and its derivatives, and it equals zero. In simpler terms, for the given exercise \(x y^{\prime \prime}+5 y^{\prime}+x y=0\), all terms contain the dependent variable \( y \) or its derivatives. This specific equation is homogeneous because there is no term independent of \( y \) or its derivatives; everything is set to zero. Such equations often allow for special methods of solution, as all terms are intrinsically linked.

Variable Coefficients

Variable coefficients in differential equations mean that the coefficients (the constants or functions multiplying the derivatives) are functions of the independent variable. In our exercise, \(x y^{\prime \prime} + 5 y^{\prime} + x y = 0\), the coefficients of \( y^{\prime \) and \( y \) are actually functions of \( x \). This often means we can't use simple methods like those for constant coefficients. Instead, special approaches like series solutions or characteristic equations become necessary.

Characteristic Equation

The characteristic equation is derived from substituting a trial solution into the differential equation. For our exercise, we use \( y(x) = x^r \). By substituting \( y = x^r, y^{\prime} = r x^{r-1}, y^{\prime\prime} = r(r-1)x^{r-2}\) into the original equation \(x y^{\prime\prime} + 5 y^{\prime} + x y = 0\):
Step-by-step, we get \[(x r(r-1)x^{r-2}) + 5(r x^{r-1}) + (x x^r) = 0\]\
Simplifying this gives \[(r(r-1) + 5r + x)x^r = 0\]. Since \(x^r eq 0\), we can divide it out, yielding the characteristic equation: \[r(r-1) + 5r + x = 0\]. To solve for \( r \), assume \( x = 0 \), giving us: \[r(r-1) + 5r = 0 <=> r^2 + 4r = 0\]. This quadratically solves to give the roots \[r = 0\] and \[r = -4\].

Series Solution

A series solution is crucial for differential equations with variable coefficients when other methods fail. This involves expressing the solution as an infinite sum of terms. For equations like \( x y^{\text{''}} + 5 y^{\text{'}} + x y = 0 \), we assume a solution in the form of \[ y(x) = \sum_{n=0}^{\text{∞}} a_n x^n\]
Series solutions can handle complexities and provide a highly accurate representation across intervals. While we didn't need to fully deploy a series solution for our exercise, understanding it broadens one's problem-solving toolkit especially for more stubborn differential equations with variable coefficients.