(a) Use the energy integral lemma to derive the family of solutions $\mathrm{y}\left(\mathrm{t}\right)=\frac{1}{\left(\mathrm{t}-\mathrm{c}\right)}$to the equation $\mathrm{y}=2{\mathrm{y}}^3$.

(b) For $\mathrm{c}\ne 0$show that these solutions are pairwise linearly independent for different values of c in an appropriate interval around t = 0.

(c) Show that none of these solutions satisfies the initial conditions $\mathrm{y}\left(0\right)=1,\mathrm{y}\text{'}\left(0\right)=2$.

The Energy Integral Lemma:

Let y(t) be a solution to the differential equation $\mathrm{y}=\mathrm{f}\left(\mathrm{y}\right)$, where f(y) is a continuous function that does not depend on y’ or the independent variable t. Let F(y) is an indefinite integral off (y), that is, $\mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{F}\left(\mathrm{y}\right)$. Then the quantity is constant; $\mathrm{E}\left(\mathrm{t}\right):=\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right)$i.e.,$\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0$.

Given that, $\mathrm{y}=2{\mathrm{y}}^3$…… (1)

To prove: the solution of the given equation is $\mathrm{y}\left(\mathrm{t}\right)=\frac{1}{\left(\mathrm{t}-\mathrm{c}\right)}$.

Let us rewrite the equation (1) into F(y) form.

$\mathrm{y}=2{\mathrm{y}}^3\Rightarrow \frac{\mathrm{d}}{\mathrm{dy}}\left(\frac{{\mathrm{y}}^4}{2}\right)$

So, $\mathrm{F}\left(\mathrm{y}\right)=\frac{{\mathrm{y}}^4}{2}$. Then,

$$\begin{gathered} \mathrm{t}=\pm \int \frac{\mathrm{dy}}{\sqrt{2\left[\mathrm{F}\left(\mathrm{y}\right)+\mathrm{k}\right]}}+\mathrm{c} \\ =\pm \int \frac{\mathrm{dy}}{\sqrt{2\left[\left(\frac{{\mathrm{y}}^4}{2}\right)+\mathrm{k}\right]}}+\mathrm{c} \end{gathered}$$

Let's put k = 0.

$$\begin{gathered} \mathrm{t}=\pm \int \frac{\mathrm{dy}}{\sqrt{\left[{\mathrm{y}}^4\right]}}+\mathrm{c} \\ =\pm \int \frac{\mathrm{dy}}{{\mathrm{y}}^2}+\mathrm{c} \\ =\pm \frac{1}{\mathrm{y}}+\mathrm{c} \end{gathered}$$

Take a positive sign. Then, we get.

$$\begin{gathered} \mathrm{t}=\frac{1}{\mathrm{y}}+\mathrm{c} \\ \mathrm{t}-\mathrm{c}=\frac{1}{\mathrm{y}} \\ \mathrm{y}=\frac{1}{\mathrm{t}-\mathrm{c}} \end{gathered}$$

Hence, $\mathrm{y}=\frac{1}{\mathrm{t}-\mathrm{c}}$is a solution.

Referring to part (a): the solution of the given equation is $\mathrm{y}=\frac{1}{\mathrm{t}-\mathrm{c}}$…… (2)

Where c is a constant.

Let us take ${\mathrm{y}}_1=\frac{1}{\mathrm{t}-{\mathrm{c}}_1}$and data-custom-editor="chemistry" ${\mathrm{y}}_2=\frac{1}{\mathrm{t}-{\mathrm{c}}_2}$around t = 0.

$$\begin{gathered} \frac{{\mathrm{y}}_1}{{\mathrm{y}}_2}=\frac{-\frac{1}{{\mathrm{c}}_1}}{-\frac{1}{{\mathrm{c}}_2}} \\ =\frac{{\mathrm{c}}_2}{{\mathrm{c}}_1} \end{gathered}$$

So, which does not give a constant. And both are linearly independent also.

Given that, $\mathrm{y}\left(0\right)=1,\mathrm{y}\text{'}\left(0\right)=2$.

Derivate the equation (2).

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

Substitute it in the solution.

$$\begin{gathered} \left(0\right)=\frac{1}{0-\mathrm{c}} \\ 1=-\frac{1}{\mathrm{c}} \\ \mathrm{c}=-1 \end{gathered}$$

Then,

$$\begin{gathered} \left(0\right)=-\frac{1}{{\left(0-\mathrm{c}\right)}^2} \\ 2=\frac{1}{{\mathrm{c}}^2} \\ {\mathrm{c}}^2=\frac{1}{2} \end{gathered}$$

Therefore, which is not true. And none of the solutions satisfies these initial conditions.