Bessel's equation - 2

[logistic_guy]

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 $\displaystyle x > 0$.

 

[logistic_guy]

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.