Question

Use the method of Problem 33 to find a second independent solution of the given equation. \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=0, \quad x>0 ; \quad y_{1}(x)=x^{-1 / 2} \sin x\)

Short answer

Question: Find a second linearly independent solution for the given differential equation \(x^2y'' + xy' + (x^2 - 0.25)y = 0\), using the method of Problem 33 and considering the known solution \(y_1(x) = x^{-\frac{1}{2}} \sin{x}\). Answer: The second linearly independent solution is given by \(y_2(x) = x^{-\frac{1}{2}} \sin{x} \cdot F(x)\) where \(F(x)\) is an antiderivative of \(x\sin^2{x}\).

Step-by-step solution

Calculate p(x)

First, calculate \(p(x)\) using \(a(x)\) and \(b(x)\): \(p(x) = \frac{b(x)}{a(x)} = \frac{x}{x^2} = \frac{1}{x}\).

Calculate the integral of p(x)

Integrate \(p(x)\) with respect to x: \(\int p(x) \, dx = \int \frac{1}{x} \, dx = \ln{x}\)

Calculate the exponential of the integral of p(x)

Calculate \(e^{-\int p(x) \, dx}\): \(e^{-\int p(x) \, dx} = e^{-\ln{x}} = x^{-1}\)

Calculate the fraction term of the formula

Calculate \(\frac{1}{y_1^2(x)}\): \(\frac{1}{y_1^2(x)} = \frac{1}{\left(x^{-\frac{1}{2}} \sin{x}\right)^2} = \frac{1}{x^{-1} \sin^2{x}} = \frac{x \sin^2{x}}{1}\)

Calculate the integral term of the formula

Calculate the integral term of the formula (we only need to calculate an antiderivative): \(\int \frac{1}{y_1^2(x)} \cdot e^{-\int p(x) \, dx} \, dx = \int x \sin^2{x} \, dx\) The result of this integral is not easily expressed in terms of elementary functions. However, the important point to note is that this integral represents an antiderivative of \(x\sin^2{x}\). Nonetheless, we can represent the result of this integral as an unknown function \(F(x)\), and express the second solution in terms of this function.

Find the second independent solution

Finally, the second linearly independent solution, \(y_2(x)\), is given by: \(y_2(x) = y_1(x) \int \frac{1}{y_1^2(x)} \cdot e^{-\int p(x) \, dx} \, dx = x^{-\frac{1}{2}} \sin{x} \cdot F(x)\) where \(F(x)\) is an antiderivative of \(x\sin^2{x}\).