Solve.
\(\displaystyle x^2y'' + xy' + \left(x^2 - \frac{1}{4}\right)y = 0\)
when you are given that \(\displaystyle y_1 = \frac{\sin x}{\sqrt{x}}\) is a solution for \(\displaystyle x > 0\).
My intuition tells me that the other solution is \(\displaystyle y_2 = \frac{\cos x}{\sqrt{x}}\). If you followed the last thread of Bessel's equation, you would notice that my intuition is correct. But I am an expert, you probably not. So, how to find the second solution when you know nothing? The good news is that there is a systematic method that helps you find a second solution when you know one solution!
First we need to write the differential equation in this form:
\(\displaystyle y'' + P(x)y' + Q(x)y = 0\)
\(\displaystyle y'' + \frac{1}{x}y' + \frac{1}{x^2}\left(x^2 - \frac{1}{4}\right)y = 0\)
This tells us that \(\displaystyle P(x) = \frac{1}{x}\) and that's all we need!
If we know one solution say \(\displaystyle y_1\), then a second solution is:
\(\displaystyle y_2 = y_1\int\frac{1}{y_1^2}e^{-\int P(x) \ dx} \ dx\)
Now we start by solving:
\(\displaystyle \int P(x) \ dx = \int \frac{1}{x} \ dx = \ln x\)
This gives us:
\(\displaystyle e^{-\ln x} = \frac{1}{x}\), then we have:
\(\displaystyle y_2 = y_1\int\frac{1}{y_1^2}\frac{1}{x} \ dx = \frac{\sin x}{\sqrt{x}}\int \frac{x}{\sin^2 x}\frac{1}{x} \ dx = \frac{\sin x}{\sqrt{x}}\int \csc^2 x \ dx\)
\(\displaystyle = -\frac{\sin x}{\sqrt{x}}\cot x = -\frac{\sin x}{\sqrt{x}}\frac{\cos x}{\sin x} = -\frac{\cos x}{\sqrt{x}}\)
While solving the integrals in this method, you may ignore the constants of integration.
Our solutions are:
\(\displaystyle y_1 = \frac{\sin x}{\sqrt{x}}\)
\(\displaystyle y_2 = -\frac{\cos x}{\sqrt{x}}\)
They can be combined together as the general solution to the original differential equation as:
\(\displaystyle y(x) = c_1\frac{\sin x}{\sqrt{x}} + c_2\frac{\cos x}{\sqrt{x}}\)
Note: We have ignored the constants of integration because the arbitrary constants \(\displaystyle c_1\) and \(\displaystyle c_2\) will take care of everything! If you looked closer, you would also notice \(\displaystyle c_2\) took care of the negative sign of the second solution.
