Question

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \(x^{2} y^{\prime \prime}+2 x y^{\prime}+4 y=0\)

Short answer

Answer: The general solution is \(y(x) = x^{\frac{-1}{2}}(c_1 \cos(\frac{\sqrt{15}}{2} \ln(x)) + c_2 \sin(\frac{\sqrt{15}}{2} \ln(x)))\), where \(c_1\) and \(c_2\) are constants.

Step-by-step solution

Identify the equation as an Euler-Cauchy equation

The given differential equation is in the form: \(x^2 y'' + 2x y' + 4 y = 0\). This matches the form of an Euler-Cauchy equation: \(x^2 y'' + a x y' + b y = 0\).

Make the substitution

To simplify the equation, let's make the substitution \(y(x) = x^r\). We will find \(y'(x)\) and \(y''(x)\) in terms of \(x\) and \(r\) to substitute back into the original equation. \(y'(x) = r x^{r-1}\) \(y''(x)= r(r-1) x^{r-2}\)

Substituting back into the original equation

Now, substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) back into the original equation and simplify. \(x^2 (r(r-1)x^{r-2}) + 2x (rx^{r-1}) + 4 (x^r) = 0\) \(x^{r} [r(r-1)+2r+4] =0\) As we are considering any interval that does not include the singular point, we can divide both sides of the equation by x^r (x ≠ 0): \(r(r-1) + 2r+4 = 0\)

Solve the characteristic equation

The characteristic equation obtained from the substitution is a quadratic equation: \(r^2 - r + 2r + 4 = 0\) \(r^2 + r + 4 = 0\) This equation has no real solutions as its discriminant \((1^2 - 4\cdot 1 \cdot 4) < 0\). Nevertheless, we can find its complex roots using the quadratic formula: \(r = \frac{-1 \pm \sqrt{1^2 - 4\cdot 1 \cdot 4}}{2} = \frac{-1 \pm \sqrt{-15}}{2}\) Thus, we have two complex roots: \(r_1 = \frac{-1 + \sqrt{-15}}{2}\) \(r_2 = \frac{-1 - \sqrt{-15}}{2}\)

Find the general solution

Given two distinct complex roots for an Euler-Cauchy equation, the general solution can be found using the following formula: \(y(x) = x^{\frac{-1}{2}}(c_1 \cos(\frac{\sqrt{15}}{2} \ln(x)) + c_2 \sin(\frac{\sqrt{15}}{2} \ln(x)))\) So, the general solution for the given differential equation that is valid in any interval not including the singular point is: \(y(x) = x^{\frac{-1}{2}}(c_1 \cos(\frac{\sqrt{15}}{2} \ln(x)) + c_2 \sin(\frac{\sqrt{15}}{2} \ln(x)))\)

Key concepts

Differential Equations

Differential equations are mathematical expressions that relate a function with its derivatives. In the context of the given exercise, we're dealing with a second-order linear homogeneous differential equation. Such an equation involves the second derivative (\( y'' \)) of the unknown function (\( y \) ) and typically represents a variety of physical phenomena, from motion to heat transfer.

For the Euler-Cauchy equation, the variable coefficients are powers of the independent variable, in this case, (\( x \) ), which makes them particularly suitable for problems with scale invariance or those describing phenomena where the rates of change are proportional to the function's current value. Understanding the form and substitution technique is a crucial step in solving these types of differential equations.

Characteristic Equation

The characteristic equation is a powerful tool for solving linear homogeneous differential equations with constant coefficients. However, when dealing with an Euler-Cauchy equation like in the given exercise, the coefficients are not constant, but we can use a transformation to derive a characteristic equation.

The substitution method transforms the original differential equation into an algebraic equation, making it much easier to handle. By assuming a solution of the form (\( y(x) = x^r \) ), the problem simplifies into finding the appropriate values of (\( r \) ) that satisfy the characteristic equation. This part of the solution process is critical as it leads to the values of (\( r \) ) that will be used to construct the general solution of the differential equation.

Complex Roots

Occasionally, the characteristic equation's solutions can be complex numbers, which is the case in our exercise. Complex roots appear in conjugate pairs due to the equation's real coefficients. This happens when the discriminant in the quadratic formula is negative.

Complex roots change our approach to finding the general solution. Instead of exponential functions, which we use for real roots, we turn to the Euler formula that connects complex exponentiation with trigonometric functions. The result is a general solution that combines trigonometric functions with the natural logarithm, reflecting the oscillatory nature of solutions when complex roots are involved. Understanding the role of complex roots in differential equations helps us interpret the behavior of solutions to physical problems and predict the oscillating patterns they may exhibit.