Chapter 7.3, Problem 23E

a.

To determine

To solve : the differential equation.

Answer to Problem 23E

The solution of differential equation is $\sin y=x^2+c$ .

Explanation of Solution

Given information :

The differential equation is $y\text{'}=2x\sqrt{1-y^2}$ .

Calculation : here the equation is separable, so write the equation in the terms of differentials and then integrate both sides,

  $\begin{array}{l}y\text{'}=2x\sqrt{1-y^2}\\ \frac{dy}{dx}=2x\sqrt{1-y^2}\text{ [}y\text{'}=\frac{dy}{dx}]\\ \frac{dy}{\sqrt{1-y^2}}=2xdx\text{ [separate the variables]}\\ {\displaystyle \int \frac{dy}{\sqrt{1-y^2}}}={\displaystyle \int 2xdx\text{ [integrate both side]}}\\ \sin y=2·\frac{x^2}{2}+c\text{ [simplify integrate] }\\ \sin y=x^2+c\text{ [simplify]}\end{array}$

Here $c$ is a arbitrary constant.

Hence, the solution of differential equation is $\sin y=x^2+c$ .

b.

To determine

To solve : the differential equation and graph the solution by using graphing calculator.

Answer to Problem 23E

The solution of differential equation is $\sin y=x^2$ .

Explanation of Solution

Given information :

The differential equation is $y\text{'}=2x\sqrt{1-y^2}$ .

  $y\left(0\right)=0$

Calculation : here the equation is separable, so write the equation in the terms of differentials and then integrate both sides,

  $\begin{array}{l}y\text{'}=2x\sqrt{1-y^2}\\ \frac{dy}{dx}=2x\sqrt{1-y^2}\text{ [}y\text{'}=\frac{dy}{dx}]\\ \frac{dy}{\sqrt{1-y^2}}=2xdx\text{ [separate the variables]}\\ {\displaystyle \int \frac{dy}{\sqrt{1-y^2}}}={\displaystyle \int 2xdx\text{ [integrate both side]}}\\ \sin y=2·\frac{x^2}{2}+c\text{ [simplify integrate] }\\ \sin y=x^2+c\text{ [simplify]}\end{array}$

Here $c$ is a arbitrary constant.

Since $y\left(0\right)=0$ , so substitute $x=0,y=0$

  $\begin{array}{l}\sin y=x^2+c\\ \sin 0={\left(0\right)}^2+c\text{ [substitute]}\\ 0=0+c\text{ [}\sin 0=0]\\ c=0\text{ [simplify]}\end{array}$

Hence, the solution of differential equation is $\sin y=x^2$ .

Using implicit plotting capability of a CAS to graph the curve $\sin y=x^2$ .

The graph is obtained as:

  

c.

To determine

To state : thatthe initial-value problem have the solution $y\left(0\right)=2$ or not.

Answer to Problem 23E

No, $y\left(0\right)=2$ is not a solution of initial value problem.

Explanation of Solution

Given information :

The differential equation is $y\text{'}=2x\sqrt{1-y^2}$ .

here the equation is separable, so write the equation in the terms of differentials and then integrate both sides,

  $\begin{array}{l}y\text{'}=2x\sqrt{1-y^2}\\ \frac{dy}{dx}=2x\sqrt{1-y^2}\text{ [}y\text{'}=\frac{dy}{dx}]\\ \frac{dy}{\sqrt{1-y^2}}=2xdx\text{ [separate the variables]}\\ {\displaystyle \int \frac{dy}{\sqrt{1-y^2}}}={\displaystyle \int 2xdx\text{ [integrate both side]}}\\ \sin y=2·\frac{x^2}{2}+c\text{ [simplify integrate] }\\ \sin y=x^2+c\text{ [simplify]}\end{array}$

Here $c$ is a arbitrary constant.

Since $y\left(0\right)=0$ , so substitute $x=0,y=0$

  $\begin{array}{l}\sin y=x^2+c\\ \sin 0={\left(0\right)}^2+c\text{ [substitute]}\\ 0=0+c\text{ [}\sin 0=0]\\ c=0\text{ [simplify]}\end{array}$

Hence, the solution of differential equation is $\sin y=x^2$ .

To check the solution substitute $x=0,y=2$ in the solution of differential equation,

  $\begin{array}{l}\sin y=x^2\\ \sin 2\ne 0\text{ [substitute]}\end{array}$

No, $y\left(0\right)=2$ is not a solution of initial value problem.