Q47 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In quantum mechanics, the study of the Schrödinger equation for the case of a harmonic oscillator leads to a consideration of Hermite's equation, $\mathrm{y}\text{'}\text{'}-2\mathrm{ty}\text{'}+\mathrm{\lambda y}=0$ , where $\mathrm{\lambda }$ is a parameter. Use the reduction of order formula to obtain an integral representation of a second linearly independent solution to Hermite's equation for the given value of $\mathrm{\lambda }$ and the corresponding solution $\mathrm{f}\left(\mathrm{t}\right)$.

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{\lambda }=4, \mathrm{f}\left(\mathrm{t}\right)=1-2{\mathrm{t}}^2 \\ \left(\mathrm{b}\right) \mathrm{\lambda }=6,\mathrm{f}\left(\mathrm{t}\right)=3\mathrm{t}-2{\mathrm{t}}^3 \end{gathered}$$

Step-by-Step Solution

Answer

1. The second linearly independent solution of the given Hermite’s equation $\mathrm{\lambda }=4, \mathrm{f}\left(\mathrm{t}\right)=1-2{\mathrm{t}}^2$ is $\mathrm{y}={\mathrm{c}}_1\left(1-2{\mathrm{t}}^2\right)+{\mathrm{c}}_2\left(1-2{\mathrm{t}}^2\right)\int \frac{{\mathrm{e}}^{{\mathrm{t}}^2}}{{\left(1-2{\mathrm{t}}^2\right)}^2}\mathrm{dt}$.
2. The second linearly independent solution of the given Hermite’s equation $\mathrm{\lambda }=6, \mathrm{f}\left(\mathrm{t}\right)=3\mathrm{t}-2{\mathrm{t}}^3$ is $\mathrm{y}={\mathrm{c}}_1\left(3\mathrm{t}-2{\mathrm{t}}^3\right)+{\mathrm{c}}_2\left(3\mathrm{t}-2{\mathrm{t}}^3\right)\int \frac{{\mathrm{e}}^{{\mathrm{t}}^2}}{{\left(3\mathrm{t}-2{\mathrm{t}}^3\right)}^2}\mathrm{dt}$.

Step 1: Finding the value of Y

Given differential equation is $\mathrm{y}\text{'}\text{'}-2\mathrm{ty}\text{'}+\mathrm{\lambda y}=0$and here $\mathrm{\lambda }=4$ and  $\mathrm{f}\left(\mathrm{t}\right)=1-2{\mathrm{t}}^2$ is one of the solutions and $\mathrm{p}\left(\mathrm{t}\right)=-2\mathrm{t}$. Then

$$\begin{gathered} {\mathrm{y}}_2={\mathrm{y}}_1\left(\mathrm{t}\right)\int \frac{{\mathrm{e}}^{-\int \mathrm{p}\left(\mathrm{t}\right)}\mathrm{dt}}{{\mathrm{y}}_1^2\left(\mathrm{t}\right)}\mathrm{dt} \\ =\left(1-2{\mathrm{t}}^2\right)\int \frac{{\mathrm{e}}^{-\int -2\mathrm{t}}\mathrm{dt}}{{\left(1-2{\mathrm{t}}^2\right)}^2}\mathrm{dt} \\ =\left(1-2{\mathrm{t}}^2\right)\int \frac{{\mathrm{e}}^{{\mathrm{t}}^2}}{{\left(1-2{\mathrm{t}}^2\right)}^2}\mathrm{dt} \end{gathered}$$

So, the solution is $\mathrm{y}={\mathrm{c}}_1\left(1-2{\mathrm{t}}^2\right)+{\mathrm{c}}_2\left(1-2{\mathrm{t}}^2\right)\int \frac{{\mathrm{e}}^{{\mathrm{t}}^2}}{{\left(1-2{\mathrm{t}}^2\right)}^2}\mathrm{dt}$ .

Step 2: Finding the value of Y

Here $\mathrm{\lambda }=6$ and $\mathrm{f}\left(\mathrm{t}\right)=1-2{\mathrm{t}}^2$ is one of the solutions and $\mathrm{p}\left(\mathrm{t}\right)=3\mathrm{t}-2{\mathrm{t}}^3$ . Then

$$\begin{gathered} {\mathrm{y}}_2={\mathrm{y}}_1\left(\mathrm{t}\right)\int \frac{{\mathrm{e}}^{-\int \mathrm{p}\left(\mathrm{t}\right)}\mathrm{dt}}{{\mathrm{y}}_1^2\left(\mathrm{t}\right)}\mathrm{dt} \\ =\left(3\mathrm{t}-2{\mathrm{t}}^3\right)\int \frac{{\mathrm{e}}^{-\int -2\mathrm{t}}\mathrm{dt}}{{\left(3\mathrm{t}-2{\mathrm{t}}^3\right)}^2}\mathrm{dt} \\ =\left(3\mathrm{t}-2{\mathrm{t}}^3\right)\int \frac{{\mathrm{e}}^{{\mathrm{t}}^2}}{{\left(3\mathrm{t}-2{\mathrm{t}}^3\right)}^2}\mathrm{dt} \end{gathered}$$

Hence, the solution is $\mathrm{y}={\mathrm{c}}_1\left(3\mathrm{t}-2{\mathrm{t}}^3\right)+{\mathrm{c}}_2\left(3\mathrm{t}-2{\mathrm{t}}^3\right)\int \frac{{\mathrm{e}}^{{\mathrm{t}}^2}}{{\left(3\mathrm{t}-2{\mathrm{t}}^3\right)}^2}\mathrm{dt}$.