Question

Determine the solution of the differential equations.

(a)$\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\text{'}}\mathbf{=}\mathit{y}$

(b) ${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\text{'}}\mathbf{-}{\mathit{x}}^{\mathbf{4}}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

(a) The solution is$y=x^{\frac{1}{2}}{Z}_1\left(2i{x}^{\frac{1}{2}}\right)$.

(b) The solution is $y=x^{\frac{1}{2}}{Z}_{\frac{1}{6}}\left(\frac{1}{3}3i{x}^3\right).$.

Step-by-step solution

Concept of Bessel functions by using equations (16.1) and (16.2):

Consider the standard form is,

${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\text{'}}\mathbf{+}\frac{\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{a}}{\mathbf{x}{\mathbf{y}}^{\mathbf{\text{'}}}}\mathbf{+}\mathbf{(}\mathit{b}\mathit{c}{\mathit{x}}^{\mathbf{c}\mathbf{-}\mathbf{1}}{\mathbf{)}}^{\mathbf{2}}\mathbf{+}\left[\frac{a^2-p^2{c}^2}{x^2}\right]\mathit{y}\mathbf{=}\mathbf{0}$

Solution to the above mentioned standard equation is,

$\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{a}}{\mathit{Z}}_{\mathbf{p}}\mathbf{\left(}\mathit{b}{\mathit{x}}^{\mathbf{c}}\mathbf{\right)}$

(a) Find the solutions of   differential equation:

Consider the differential equation as below.

$$\begin{gathered} x{y}^{\text{'}}\text{'}=y\text{'}\text{'} \\ {y}^{\text{'}}\text{'}\text{'}-\frac{y}{X}=0 \end{gathered}$$

Compare the equation with standard Bessel function equation as follows:

$$\begin{gathered} a=\frac{1}{2} \\ c=\frac{1}{2} \\ p=1 \\ b=2i \end{gathered}$$

So,

role="math" localid="1659268440330" $y=x^{\frac{1}{2}}{Z}_1\left(2i{x}^{\frac{1}{2}}\right)$

Now,

$$\begin{gathered} y=x^{\frac{1}{2}}{Z}_1\left(2i{x}^{\frac{1}{2}}\right) \\ y=x^{\frac{1}{2}}{I}_1\left(2i{x}^{\frac{1}{2}}\right) \\ (\because I_1=i^{(}-1)I_1\left(ix\right)) \end{gathered}$$

(b) Find the solutions of y''-x4y=0  differential equation:

Consider the differential equation as below.

$y^{\text{'}}\text{'}-x^{4}y=0$

Then, $y^{\text{'}}\text{'}-\frac{y}{X}=0$.

Compare the equation with standard Bessel function equation as follows:

$$\begin{gathered} a=\frac{1}{2} \\ c=3 \\ p=\frac{1}{6} \\ b=\frac{1}{3}i \end{gathered}$$

So,

$y=x^{\frac{1}{2}}{Z}_{\frac{1}{6}}\left(\frac{1}{3}3i{x}^3\right).$

Now,

$$\begin{gathered} y=x^{\frac{1}{2}}{Z}_{\frac{1}{6}}\left(\frac{1}{3}3i{x}^3\right) \\ y=x^{\frac{1}{2}}{I}_{\frac{1}{6}}\left(\frac{1}{3}{x}^3\right) \\ \left(\because I_1=i^{-1}{I}_1\left(ix\right)\right) \end{gathered}$$