Chapter 7.1, Problem 8E

a.

To determine

To comment : On the graph of the solution of the $y^{\prime }=x{y}^3$ when $x$ is close to $0$ and when $x$ is large.

Answer to Problem 8E

When $x$ is close to $0$ , the slope of the function is close to $0$ .

When $x$ become large then the slope of the function becomes very large in magnitude.

Explanation of Solution

Given information :

  $y^{\prime }=x{y}^3$

Formula used :

  ${\displaystyle \int x^n}dn=\frac{x^{n+1}}{n+1}+c$

  $x\ge y\Rightarrow -x\le -y$ and $x\ge y\Rightarrow \frac{1}{x}\le \frac{1}{y}$

Calculation :

  $\begin{array}{l}{y}^{\prime }=x{y}^3\\ \Rightarrow \frac{dy}{y^3}=xdx\end{array}$

Integrating we get,

  $\begin{array}{l}\frac{y^{-2}}{-2}+\frac{c}{2}=\frac{x^2}{2}\\ \Rightarrow -\frac{1}{y^2}=x^2-c\\ \Rightarrow y^2=\frac{1}{\left(c-x^2\right)}\mathrm{...........}\left(i\right)\\ \Rightarrow y=\pm \frac{1}{{\left(c-x^2\right)}^{\frac{1}{2}}}\\ \Rightarrow y^3=\pm \frac{1}{{\left(c-x^2\right)}^{\frac{3}{2}}}\\ \Rightarrow x{y}^3=\pm \frac{x}{{\left(c-x^2\right)}^{\frac{3}{2}}}\end{array}$

  $\therefore y^{\prime }=\pm \frac{x}{{\left(c-x^2\right)}^{\frac{3}{2}}}$

Now, $\underset{x\to 0}{\lim }{y}^{\prime }=0$

i.e. When $x$ is close to $0$ , the slope of the function is close to $0$ .

  $y^2=\frac{1}{\left(c-x^2\right)}$ [ equation (i)]

  $\begin{array}{l}\Rightarrow \frac{1}{c-x^2}\ge 0\text{ as }{y}^2\text{ can never be negative}\\ \Rightarrow c-x^2\le 0\Rightarrow x^2-c\ge 0\\ \Rightarrow x^2\ge c\\ \Rightarrow -\sqrt{c}<x<\sqrt{c}\end{array}$

Therefore, when we say $x$ is large , it means $x$ is close to $\sqrt{c}$ from left.

  $\therefore \underset{x\to \sqrt{c}}{\lim }{y}^{\prime }=\underset{x\to \sqrt{c}}{\lim }\pm \frac{x}{{\left(c-x^2\right)}^{\frac{3}{2}}}=\pm \infty$

i.e. When $x$ become large then the slope of the function becomes very large in magnitude.

b.

To determine

To verify : All the members of the family of the function $y={\left(c-x^2\right)}^{-\frac{1}{2}}$ is the solution of the given differential equation.

Explanation of Solution

Given information :

Differential equation: $y^{\prime }=x{y}^3$

  $y={\left(c-x^2\right)}^{-\frac{1}{2}}$

Formula used :

  $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

  $\frac{d}{dx}\left(u\pm v\right)=\frac{d}{dx}u\pm \frac{d}{dx}v$

Calculation :

  $y={\left(c-x^2\right)}^{-\frac{1}{2}}$

Differentiating both sides with respect to $x$ , we get

  $\begin{array}{c}{y}^{\prime }=-\frac{1}{2}{\left(c-x^2\right)}^{-\frac{1}{2}-1}\times \left(-2x\right)\\ =\frac{x}{{\left(c-x^2\right)}^{\frac{3}{2}}}\\ =x{\left[{\left(c-x^2\right)}^{-\frac{1}{2}}\right]}^3\end{array}$

  $\Rightarrow y^{\prime }=x{y}^3$

So, $y={\left(c-x^2\right)}^{-\frac{1}{2}}$ is a solution of differential equation $y^{\prime }=x{y}^3$ .

c.

To determine

To graph : The several member of the family of the solution on the common screen. And identify whether the graph confirm the conclusion of the part (a) or not.

Answer to Problem 8E

  

Yes, the graphs confirm the conclusion of part (a).

Explanation of Solution

Given information :

  $y={\left(c-x^2\right)}^{-\frac{1}{2}}$

Graph :

  

Interpretation :

The graphs are drawn for different value of $c$ .

  $C_1=1;C_2=2;C_3=3;C_4=4;=C_5=9$

Yes, the graphs confirm the conclusion of part (a).

d.

To determine

To find : A solution of a initial value problem.

Answer to Problem 8E

  $y=\pm \frac{2}{{\left(1-4x^2\right)}^{\frac{1}{2}}}$ is the particular solution.

Explanation of Solution

Given information :

Differential equation: $y^{\prime }=x{y}^3$

  $y\left(0\right)=2$

  $y=\pm \frac{1}{{\left(c-x^2\right)}^{\frac{1}{2}}}$

Concept used :

If $y=f\left(x\right)$ is a solution of a differential equation then it must satisfy the equation.

Calculation :

We have in part (b) $y=\pm \frac{1}{{\left(c-x^2\right)}^{\frac{1}{2}}}$ is the family of solution of the differential equation $y^{\prime }=x{y}^3$

Now, $x=0$ ; $y=2$

Then, $2=\pm \frac{1}{c^{\frac{1}{2}}}\Rightarrow c=\frac{1}{4}$

  $y=\pm \frac{1}{{\left(\frac{1}{4}-x^2\right)}^{\frac{1}{2}}}$

i.e. $y=\pm \frac{2}{{\left(1-4x^2\right)}^{\frac{1}{2}}}$ is the particular solution.