Chapter 7.1, Problem 7E

a.

To determine

To identify : Nature of the solution of differential equation $y^{\prime }=-y^2$ without solving.

Answer to Problem 7E

The solution will be non increasing.

Explanation of Solution

Given information :

  $y^{\prime }=-y^2$

As $y^{\prime }=-y^2\le 0$ then we can easily say that the solution is non increasing.

b.

To determine

To verify : All the members of the function $y=\frac{1}{x+c}$ are the solution of the differential equation $y^{\prime }=-y^2$

Explanation of Solution

Given information :

Differential equation: $y^{\prime }=-y^2$

  $y=\frac{1}{x+c}$

Concept used :

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

Calculation :

  $y=\frac{1}{x+c}$

Differentiating with respect to $x$ , we get,

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

  $\Rightarrow y^{\prime }=-y^2$

i.e. the family, $y=\frac{1}{x+c}$ is a solution of the differential equation $y^{\prime }=-y^2$

c.

To determine

To find : The solution of given differential equation $y^{\prime }=-y^2$ other than part (b).

Answer to Problem 7E

  $y=0$

Explanation of Solution

Given information :

Differential equation: $y^{\prime }=-y^2$

Solution of the differential equation at part (b) : $y=\frac{1}{x+c}$

Formula used :

Trivial solution

Calculation :

We, consider $y$ as $f\left(x\right)$ .

Therefore, $f^{\prime }\left(x\right)=-{\left[f\left(x\right)\right]}^2$

Now, $f\left(x\right)=0$ clearly does solve the equation. It does so trivially that there is no need to ask why. You need only understand what the equation asserts.

d.

To determine

To find : The solution of the initial value problem.

Answer to Problem 7E

The particular solution is $y=\frac{1}{x+2}$

Explanation of Solution

Given information :

  $y^{\prime }=-y^2$ ; $y=\frac{1}{x+c}$ ; $y\left(0\right)=0.5$

Concept used :

We put $x=0$ , $y=0.5$ at $y$ to calculate particular solution.

Calculation :

  $y=\frac{1}{x+c}$

At $x=0$ , $y=0.5$

Therefore, $0.5=\frac{1}{c}\Rightarrow c=2$

Therefore, the particular solution is $y=\frac{1}{x+2}$