The Bessel equation of order one-half ${\mathbf{t}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}\left({\mathbf{t}}^{\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\right)\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{t}\mathbf{>}\mathbf{0}$has two linearly independent solutions,${\mathbf{y}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{cost}\mathbf{,}{\mathbf{y}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{sin}$.

Find a general solution to the non-homogeneous equation.

${\mathbf{t}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}\left({\mathbf{t}}^{\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\right)\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{5}\mathbf{/}\mathbf{2}}\mathbf{,}\mathbf{t}\mathbf{>}\mathbf{0}$

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To find a general solution to the given equation, first, we need to find a particular solution.

We will do that by using the method of variation of parameters and assume that a particular solution has a form of ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{v}}_1\left(\mathrm{t}\right){\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{v}}_2\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)$, where${\mathrm{y}}_1\left(\mathrm{t}\right)={\mathrm{t}}^{-1/2}\mathrm{cost}$ and${\mathrm{y}}_2\left(\mathrm{t}\right)={\mathrm{t}}^{-1/2}\mathrm{sint}$ are two linearly independent solutions given in the problem.

Before we proceed, we need to transform the given equation so that a coefficient multiplying $\mathrm{y}\text{'}\text{'}$is 1. To obtain that we will divide the given equation by ${\mathrm{t}}^2$:

$\mathrm{y}\text{'}\text{'}+\frac{1}{\mathrm{t}}\mathrm{y}\text{'}+\left(1-\frac{1}{4{\mathrm{t}}^2}\right)\mathrm{y}={\mathrm{t}}^{1/2}$

One can find the functions and from

$$\begin{gathered} {\mathrm{v}}_1\left(\mathrm{t}\right)=\int \frac{-\mathrm{g}\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)}{\mathrm{W}\left[{\mathrm{y}}_1\left(\mathrm{t}\right),{\mathrm{y}}_2\left(\mathrm{t}\right)\right]}\mathrm{dt}, \\ {\mathrm{v}}_2\left(\mathrm{t}\right)=\int \frac{\mathrm{g}\left(\mathrm{t}\right){\mathrm{y}}_1\left(\mathrm{t}\right)}{\mathrm{W}\left[{\mathrm{y}}_1\left(\mathrm{t}\right),{\mathrm{y}}_2\left(\mathrm{t}\right)\right]}\mathrm{dt} \end{gathered}$$

Where $\mathrm{g}\left(\mathrm{t}\right)={\mathrm{t}}^{1/2}$is a non-homogeneous part of the equation and $\mathrm{W}\left[{\mathrm{y}}_1\left(\mathrm{t}\right),{\mathrm{y}}_2\left(\mathrm{t}\right)\right]$is the Wronskian of the functions ${\mathrm{y}}_1\left(\mathrm{t}\right)$and ${\mathrm{y}}_2\left(\mathrm{t}\right)$

One can find the functions ${\mathrm{v}}_1\left(\mathrm{t}\right)$and ${\mathrm{v}}_2\left(\mathrm{t}\right)$

One can take ${\mathrm{C}}_1={\mathrm{C}}_2=0$.

Now one has that the particular solution is:

$$\begin{gathered} {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{v}}_1\left(\mathrm{t}\right){\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{v}}_2\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right) \\ =(\mathrm{tcost}-\mathrm{sint})\times {\mathrm{t}}^{-1/2}\mathrm{cost}+(\mathrm{tsint}+\mathrm{cost})\times {\mathrm{t}}^{-1/2}\mathrm{sint} \\ ={\mathrm{t}}^{1/2}{\cos }^2\mathrm{t}-{\mathrm{t}}^{-1/2}\mathrm{sintcost}+{\mathrm{t}}^{1/2}{\sin }^2\mathrm{t}+{\mathrm{t}}^{-1/2}\mathrm{costsint} \\ ={\mathrm{t}}^{1/2} \end{gathered}$$

Therefore, the general solution to the given differential equation is:

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{t}}^{-1/2}\mathrm{cost}+{\mathrm{c}}_2{\mathrm{t}}^{-1/2}\mathrm{sint}+{\mathrm{t}}^{1/2}$

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