Consider the linear equation ${\mathbf{t}}^{\mathbf{2}}\mathbf{y}\text{'}\text{'}\mathbf{-}\mathbf{3}\mathbf{ty}\text{'}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}$for$\mathbf{-}\mathbf{\infty }\mathbf{<}\mathbf{t}\mathbf{<}\mathbf{\infty }$

(a). Verify that ${\mathbf{y}}_{\mathbf{1}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{t}$and ${\mathbf{y}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}$are two solutions to 21on $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$. Furthermore, show that ${\mathbf{y}}_{\mathbf{1}}\left({\mathbf{t}}_{\mathbf{0}}\right){\mathbf{y}}_{\mathbf{2}}^{}\text{'}\left({\mathbf{t}}_{\mathbf{0}}\right){\mathbf{-}\mathbf{y}}_{\mathbf{1}}^{}\text{'}\left({\mathbf{t}}_{\mathbf{0}}\right){\mathbf{y}}_{\mathbf{2}}\left({\mathbf{t}}_{\mathbf{0}}\right)\ne \mathbf{0}\mathbf{,}{\mathbf{t}}_{\mathbf{0}}\mathbf{=}\mathbf{1}$.

(b). Prove that ${\mathbf{y}}_1\left(\mathbf{t}\right)$and ${\mathbf{y}}_{\mathbf{2}}\left(\mathbf{t}\right)$are linearly independent on $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$.

(c). Verify that the function ${\mathbf{y}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{|}\mathbf{t}{\mathbf{|}}^{\mathbf{3}}$is also a solution to 21on $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$.

(d). Prove that there is no choice of constants ${\mathbf{c}}_{\mathbf{1}}\mathbf{,}{\mathbf{c}}_{\mathbf{2}}$such that ${\mathbf{y}}_{\mathbf{3}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{y}}_{\mathbf{1}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{y}}_{\mathbf{2}}\left(\mathbf{t}\right)$for all tin $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$. [Hint: Argue that the contrary assumption leads to a contradiction.]

(e). From parts (c)and (d), we see that there is at least one solution to 21on $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$that is not expressible as a linear combination of the solutions ${\mathbf{y}}_{\mathbf{1}}\left(\mathbf{t}\right)\mathbf{,}{\mathbf{y}}_{\mathbf{2}}\left(\mathbf{t}\right)$. Does this provide a counterexample to the theory in this section? Explain.

Let's show that${\mathrm{y}}_1\left(\mathrm{t}\right)=\mathrm{t}$is a solution to the given equation. First, we will find the required derivatives:${\mathrm{y}}_1^{}\text{'}\left(\mathrm{t}\right)=1,{\mathrm{y}}_1^{}\text{'}\text{'}\left(\mathrm{t}\right)=0.$

Substituting this into the given equation we have that and since${\mathrm{y}}_1\left(\mathrm{t}\right)$satisfies the given equation it is a solution.

One will repeat the steps for ${\mathrm{y}}_2\left(\mathrm{t}\right)={\mathrm{t}}^3$:

${\mathrm{y}}_2^{}\text{'}\left(\mathrm{t}\right)=3{\mathrm{t}}^2,{\mathrm{y}}_2^{}\text{'}\text{'}\left(\mathrm{t}\right)=6\mathrm{t}\iff {\mathrm{t}}^2{\mathrm{y}}_2^{}\text{'}\text{'}{\left(\mathrm{t}\right)-3\mathrm{ty}}_2^{}\text{'}\left(\mathrm{t}\right)+3{\mathrm{y}}_2\left(\mathrm{t}\right)={\mathrm{t}}^2\times 6\mathrm{t}-3\mathrm{t}\times 3{\mathrm{t}}^2+3{\mathrm{t}}^3=9{\mathrm{t}}^3-9{\mathrm{t}}^3\equiv 0$

And again${\mathrm{y}}_2\left(\mathrm{t}\right)$ satisfies the given equation and therefore it is a solution.

Now one has to show that${\mathrm{y}}_1\left({\mathrm{t}}_0\right){\mathrm{y}}_2^{}\text{'}\left({\mathrm{t}}_0\right){-\mathrm{y}}_1^{}\text{'}\left({\mathrm{t}}_0\right){\mathrm{y}}_2\left({\mathrm{t}}_0\right)\ne 0$ for $\mathrm{t}=1$. We have already found${\mathrm{y}}_1^{}\text{'}\left(\mathrm{t}\right)$ and ${\mathrm{y}}_2^{}\text{'}\left(\mathrm{t}\right)$, so one has that

$$\begin{gathered} {\mathrm{y}}_1\left(\mathrm{t}\right){\mathrm{y}}_2\text{'}\left(\mathrm{t}\right)-{\mathrm{y}}_1\text{'}\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)=\mathrm{t}\times 3{\mathrm{t}}^2-1\times {\mathrm{t}}^3 \\ =3{\mathrm{t}}^3-{\mathrm{t}}^3 \\ =2{\mathrm{t}}^3 \\ {\mathrm{y}}_1\left(1\right){\mathrm{y}}_2\text{'}\left(1\right)-{\mathrm{y}}_1\text{'}\left(1\right){\mathrm{y}}_2\left(1\right)=2\times 1^3 \\ =2\times 1 \\ =2 \\ \ne 0. \end{gathered}$$

One knows from the Lemma 3 that if the Wronskian of two functions is equal to 0, then those functions are linearly dependent. But it is easy to show that if two functions are linearly dependent, then their Wronskian must be 0 at any point $\mathrm{t}\in (-\infty ,\infty )$.

Assume that two functions ${\mathrm{f}}_1\left(\mathrm{t}\right)$and ${\mathrm{f}}_2\left(\mathrm{t}\right)$are linearly dependent, i.e., there is some constant c such that ${\mathrm{f}}_1\left(\mathrm{t}\right)={\mathrm{cf}}_2\left(\mathrm{t}\right)$. Let's find their Wronskian:

Now one has that the Wronskian for two functions are 0 at any point 1 if and only if those functions are linearly dependent. But we have shown in (a) that the Wronskian for${\mathrm{y}}_1\left(\mathrm{t}\right)$and${\mathrm{y}}_2\left(\mathrm{t}\right)$at the point$\mathrm{t}=1$are$\mathrm{W}\left[{\mathrm{y}}_1\left(1\right),{\mathrm{y}}_2\left(1\right)\right]=2\ne 0$so we can conclude that those two functions are linearly independent.

Notice that ${\mathrm{y}}_3\left(\mathrm{t}\right)=|\mathrm{t}{|}^3$can be transformed to ${\mathrm{y}}_3\left(\mathrm{t}\right)=(\mathrm{sgnt}\times \mathrm{t}{)}^3=(\mathrm{sgnt}{)}^3{\mathrm{t}}^3$, where $\mathrm{sgnt}=1$when $\mathrm{t}>0,\mathrm{sgnt}=-1$,when$\mathrm{t}<0$ and$\mathrm{sgnt}=0$ if $\mathrm{t}=0$. So, we can treat as a constant in the function ${\mathrm{y}}_3\left(\mathrm{t}\right)$, and then the required derivatives are

$$\begin{gathered} {\mathrm{y}}_3^{}\text{'}\left(\mathrm{t}\right)=(\mathrm{sgnt}{)}^3\times 3{\mathrm{t}}^2 \\ =3(\mathrm{sgnt}{)}^3{\mathrm{t}}^2,{\mathrm{y}}_3^{}\text{'}\text{'}\left(\mathrm{t}\right) \\ =3(\mathrm{sgnt}{)}^3\times 2\mathrm{t} \\ =6(\mathrm{sgnt}{)}^3\mathrm{t} \end{gathered}$$

Substituting this into the given equation we have that

$$\begin{gathered} {\mathrm{t}}^2{\mathrm{y}}_3^{}\text{'}\text{'}\text{'}{\left(\mathrm{t}\right)-3\mathrm{ty}}_3^{}\text{'}\left(\mathrm{t}\right)+3\mathrm{y}\left(\mathrm{t}\right)={\mathrm{t}}^2\times 6(\mathrm{sgnt}{)}^3\mathrm{t}-3\mathrm{t}\times 3(\mathrm{sgnt}{)}^3{\mathrm{t}}^2+3(\mathrm{sgnt}{)}^3{\mathrm{t}}^3 \\ =9(\mathrm{sgnt}{)}^3{\mathrm{t}}^3-9(\mathrm{sgnt}{)}^3{\mathrm{t}}^2 \\ \equiv 0 \end{gathered}$$

Hence${\mathrm{y}}_3\left(\mathrm{t}\right)$ satisfies the given equation, so it is a solution.

Assume that there are some constants${\mathrm{c}}_1$ and${\mathrm{c}}_2$ such that$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{y}}_2\left(\mathrm{t}\right)$

Let's take $\mathrm{t}=1$and $\mathrm{t}=-1$and try to find those two constants:

$$\begin{gathered} 1=\left|1\right| \\ ={\mathrm{c}}_1\times 1+{\mathrm{c}}_2\times 1^3 \\ ={\mathrm{c}}_1+{\mathrm{c}}_2 \\ 1=|-1| \\ ={\mathrm{c}}_1\times (-1)+{\mathrm{c}}_2\times (-1{)}^3 \\ =-{\mathrm{c}}_1-{\mathrm{c}}_2 \end{gathered}$$

This system has a unique solution if the determinant of the system is not equal to 0, but in our case the determinant of the system is:

$\begin{array}{rcl}D& =& \left|\begin{array}{cc}1& 1\\ -1& -1\end{array}\right|\\ & =& -1+1\\ & =& 0\end{array}$

So, this system has no solution and therefore there is no choice of constants ${\mathrm{c}}_1$and ${\mathrm{c}}_2$such that.

${\mathrm{y}}_3\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{y}}_2\left(\mathrm{t}\right)$

No, because the theory in this section is valid for equations in the standard form and we didn't use that form. Furthermore, this equation cannot be expressed in the standard form because if we want to express the given equation in standard form, we would have to divide by ${\mathrm{t}}^2$, but we cannot do that for $\mathrm{t}=0$.