Question

(a) Determine two singular solutions of the DE in Problem 10.

(b) If the initial condition y(5) = 0is as prescribed in Problem 10, then what is the largest interval I over which the solution is defined? Use a graphing utility to graph the solution curve for the IVP.

Short answer

1. Therefore, the two singular solutions are y = 2x and y = 3x
2. The result is$y=x\sin \left(In\frac{x}{5}\right),I=\left(0,\infty \right)$

Step-by-step solution

Given Information

$$\begin{gathered} y\text{'}=\frac{y}{x}+\frac{1}{x}.\sqrt{x^2\left(1-\frac{y^2}{x^2}\right)} \\ =\frac{y}{x}+\frac{x}{x}.\sqrt{1-\frac{y^2}{x^2}} \\ =\frac{y}{x}+\sqrt{1-{\left(\frac{y}{x}\right)}^2}---------->\left(1\right) \\ Letu=\frac{y}{x}\Rightarrow y=ux,then \\ y\text{'}=u\text{'}x+u-----------\left(2\right) \end{gathered}$$

Substitute the equation

Substitute y = ux and (2) in (1).

$$\begin{gathered} u\text{'}x+u=u+\sqrt{1-u^2} \\ u\text{'}x=\cancel{u}-\cancel{u}+\sqrt{1-u^2} \end{gathered}$$

Divide by x

$$\begin{gathered} u\text{'}x=\sqrt{1-u^2} \\ u\text{'}=\frac{\sqrt{1-u^2}}{x} \\ \frac{du}{dx}=\frac{\sqrt{1-u^2}}{x} \end{gathered}$$

Separate variables

$\frac{du}{\sqrt{1-u^2}}=\frac{dx}{x}$

Integrate each side

$$\begin{gathered} \int \frac{du}{\sqrt{1-u^2}}=\int \frac{dx}{x} \\ {\sin }^{-1}u=In\left|x\right|+c \\ \left(u=\frac{y}{x}\right) \\ {\sin }^{-1}\left(\frac{y}{x}\right)=In\left|x\right|+c \\ Multiplybyx \\ \frac{y}{x}=\sin \left(In\left|x\right|+c\right) \\ y=x\sin \left(In\left|x\right|+c\right)---------->\left(3\right) \end{gathered}$$

Substitute the value in (3) equation

Put y = 2x in (3).

Divide by x

$$\begin{gathered} 2x=x\sin \left(In\left|x\right|+c\right) \\ 2=\sin \left(In\left|x\right|+c\right) \end{gathered}$$

This is not possible, because the value of sine function never equals to 2.

Therefore, one singular solution is y = 2x.

Put y = 3x in (3)

Divide by x

role="math" $\begin{array}{rcl}3x& =& x\sin \left(In\left|x\right|+c\right)\\ 3& =& \sin \left(In\left|x\right|+c\right)\end{array}$

This is not possible, because the value of sine function never equals to 3

Therefore, another singular solution is y = 3x

(a) So, the two singular solutions are y = 2x and y = 3x

Use the initial condition y(5) = 0 to get c.

$$\begin{gathered} 0=5\sin \left(In\left|5\right|+c\right) \\ 0=\sin \left(In\left|5\right|+c\right) \\ {\sin }^{-1}\left(0\right)=In\left|5\right|+c \\ c=-In\left|5\right| \end{gathered}$$

Substitute the value in the equation

Substitute the value $c=-In\left|5\right|$in the equation (3)

$$\begin{gathered} \left(Ina-Inb=In\frac{a}{b}\right) \\ y=x\sin \left(In\left|x\right|-In\left|5\right|\right) \\ y=x\sin \left(In\frac{x}{5}\right) \end{gathered}$$

Largest interval is $\left(0,\infty \right)$

The solution curve is shown in the graph below: