Show that the three solutions $\frac{1}{{\left(1-\mathrm{t}\right)}^2},\frac{1}{{\left(2-\mathrm{t}\right)}^2},$and$\frac{1}{{\left(3-\mathrm{t}\right)}^2}$to localid="1664027547373" ${\mathrm{y}}^{\text{'}\text{'}}-6{\mathrm{y}}^2=0$are linearly independent on (-1, 1). (See Problem 35, Exercises 4.2, page 164.)

Linear Dependence of Three Functions: Three functions${\mathrm{y}}_1\left(\mathrm{t}\right),{\mathrm{y}}_2\left(\mathrm{t}\right),$ and ${\mathrm{y}}_3\left(\mathrm{t}\right)$are said to be linearly dependent on an interval I if, on I, at least one of these functions is a linear combination of the remaining two [e.g., if] ${\mathrm{y}}_1\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{y}}_2\left(\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{y}}_3\left(\mathrm{t}\right)$.

Equivalent (compare Problem 33), ${\mathrm{y}}_1\left(\mathrm{t}\right),{\mathrm{y}}_2\left(\mathrm{t}\right),$and ${\mathrm{y}}_3\left(\mathrm{t}\right)$are linearly dependent on I if there exist constants ${\mathrm{C}}_1,{\mathrm{C}}_2,$and ${\mathrm{C}}_3$not all zero, such that${\mathrm{C}}_1{\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{C}}_2{\mathrm{y}}_2\left(\mathrm{t}\right)+{\mathrm{C}}_3{\mathrm{y}}_3\left(\mathrm{t}\right)=0$ for all t in I.

otherwise, we say that these functions are linearly independent on I.

Given that, $\mathrm{y}-6{\mathrm{y}}^2=0$........(1)

To prove: $\frac{1}{{\left(1-\mathrm{t}\right)}^2},\frac{1}{{\left(2-\mathrm{t}\right)}^2}$and $\frac{1}{{\left(3-\mathrm{t}\right)}^2}$are the three solutions of the given equation.

Let us check the solutions for different cases.

Case (1):

If $\mathrm{y}=\frac{1}{{\left(1-\mathrm{t}\right)}^2}$. Then, find its first and second-order derivatives.

$$\begin{gathered} \mathrm{y}\text{'}=\frac{2}{{\left(1-\mathrm{t}\right)}^3} \\ \mathrm{y}=\frac{6}{{\left(1-\mathrm{t}\right)}^4} \end{gathered}$$

Substitute the derivatives in equation (1).

$$\begin{gathered} \mathrm{y}-6{\mathrm{y}}^2=\frac{6}{{\left(1-\mathrm{t}\right)}^4}-6{\left(\frac{1}{{\left(1-\mathrm{t}\right)}^2}\right)}^2 \\ =\frac{6}{{\left(1-\mathrm{t}\right)}^4}-\frac{6}{{\left(1-\mathrm{t}\right)}^4} \\ =0 \end{gathered}$$

Therefore, $\mathrm{y}=\frac{1}{{\left(1-\mathrm{t}\right)}^2}$is a solution.

Case (2):

If $\mathrm{y}=\frac{1}{{\left(2-\mathrm{t}\right)}^2}$. Then, find its first and second-order derivatives.

$$\begin{gathered} \mathrm{y}\text{'}=\frac{2}{{\left(2-\mathrm{t}\right)}^3} \\ \mathrm{y}=\frac{6}{{\left(2-\mathrm{t}\right)}^4} \end{gathered}$$

Substitute the derivatives in equation (1).

data-custom-editor="chemistry" $$\begin{gathered} \mathrm{y}-6{\mathrm{y}}^2=\frac{6}{{\left(2-\mathrm{t}\right)}^4}-6{\left(\frac{1}{{\left(2-\mathrm{t}\right)}^2}\right)}^2 \\ =\frac{6}{{\left(2-\mathrm{t}\right)}^4}-\frac{6}{{\left(2-\mathrm{t}\right)}^4} \\ =0 \end{gathered}$$

Hence, is a solution.

Hence, $\mathrm{y}=\frac{1}{{\left(2-\mathrm{t}\right)}^2}$is a solution.

If $\mathrm{y}=\frac{1}{{\left(3-\mathrm{t}\right)}^2}$. Then, find its first and second-order derivatives.

$$\begin{gathered} \mathrm{y}\text{'}=\frac{2}{{\left(3-\mathrm{t}\right)}^3} \\ \mathrm{y}=\frac{6}{{\left(3-\mathrm{t}\right)}^4} \end{gathered}$$

Substitute the derivatives in equation (1).

$$\begin{gathered} \mathrm{y}-6{\mathrm{y}}^2=\frac{6}{{\left(3-\mathrm{t}\right)}^4}-6{\left(\frac{1}{{\left(3-\mathrm{t}\right)}^2}\right)}^2 \\ =\frac{6}{{\left(3-\mathrm{t}\right)}^4}-\frac{6}{{\left(3-\mathrm{t}\right)}^4} \\ =0 \end{gathered}$$

Since, $\mathrm{y}=\frac{1}{{\left(3-\mathrm{t}\right)}^2}$is a solution. So, all the three are solutions of the given equation.

Now check the linear independence,

${\mathrm{c}}_1\left(\frac{1}{{\left(1-\mathrm{t}\right)}^2}\right)+{\mathrm{c}}_2\left(\frac{1}{{\left(2-\mathrm{t}\right)}^2}\right)+{\mathrm{c}}_3\left(\frac{1}{{\left(3-\mathrm{t}\right)}^2}\right)=0$

The above equation is zero if and only if ${\mathrm{c}}_1={\mathrm{c}}_2={\mathrm{c}}_3=0$. That is, all the three solutions are linear independent also.