Question

Find the general solution of the given Euler equation on \((0, \infty)\). $$ x^{2} y^{\prime \prime}-3 x y^{\prime}+13 y=0 $$

Short answer

Short Answer: The general solution of the given Euler equation on the interval (0, ∞) is: $$ y(x) = x^2(c_1 \cos(3 \ln x) + c_2 \sin(3 \ln x)) $$ where \(c_1\) and \(c_2\) are arbitrary constants.

Step-by-step solution

Perform a substitution to transform the Euler equation

Let's perform the substitution \(y(x) = x^r\). Then, find the first derivative: \(y'(x) = r \cdot x^{r - 1}\), and the second derivative: \(y''(x) = r (r - 1) x^{r - 2}\). Substitute these expressions into the Euler equation: $$ x^{2} (r (r - 1) x^{r - 2}) - 3 x (r \cdot x^{r - 1}) + 13 (x^r) = 0 $$

Solve the transformed equation

Simplify the equation and collect the terms: $$ r(r - 1)x^r - 3rx^r + 13 x^r = 0 $$ Now, factor out the common term \(x^r\): $$ x^r(r (r - 1) - 3r + 13) = 0 $$ Since \(x > 0\), we can divide by \(x^r\) without altering the equation. This leaves us with a quadratic equation in \(r\): $$ r(r - 1) - 3r + 13 = 0 $$ or simply: $$ r^2 - 4r+ 13 = 0 $$ Using the quadratic formula, we can find the roots of the equation: $$ r = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(13)}}{2(1)} = 2 \pm 3i $$ Since the roots are complex conjugates, the general solution will be the linear combination of the real and imaginary parts: $$ y(x) = x^2 (c_1 \cos(3 \ln x) + c_2 \sin(3 \ln x)) $$

Write the general solution

The general solution of the Euler equation on \((0, \infty)\) is: $$ y(x) = x^2 (c_1 \cos(3 \ln x) + c_2 \sin(3 \ln x)) $$ where \(c_1\) and \(c_2\) are arbitrary constants.