Question

(a) By inspection and a one-parameter family of solutions of the differential equation $\mathbf{xy}\text{'}=\mathbf{y}$. Verify that each member of the family is a solution of the initial-value problem $\mathbf{xy}\text{'}=\mathbf{y}$, $\mathbf{y}\left(\mathbf{0}\right)=\mathbf{0}$.

(b) Explain part (a) by determining a region R in the xy-plane for which the differential equation $\mathbf{xy}\text{'}=\mathbf{y}$ would have a unique solution through a point $\left({\mathbf{x}}_{\mathbf{0}},{\mathbf{y}}_{\mathbf{0}}\right)$ in R.

(c) Verify that the piecewise-defined function

$\mathit{y}\mathbf{=}\{\begin{array}{l}0,x<0\\ x,x\ge 0\end{array}$ satisfies the condition $\mathbf{y}\left(\mathbf{0}\right)=\mathbf{0}$. Determine whether this function is also a solution of the initial-value problem in part (a).

Short answer

a) $y=cx$

b) $x>0$or $x<0$ we see that $f$ and $\frac{df}{dy}$ are not defined at $x=0$

c) Not a solution to the initial value problem in part (a)

Step-by-step solution

Existence of a Unique solution

Let R be a rectangular region in the xy-plane defined by $a⩽x⩽b,c⩽y⩽d$ that contains the point $(x_0,y_0)$in its interior. If $f(x,y)$ and $\partial f/\partial y$are continuous on R, then there exists some interval $I_0:(x_0-h,x_0+h),h>0$ contained in $[a,b]$ and a unique function $y\left(x\right)$, defined on $I_0$, that is a solution of the initial-value problem (2).

2 (a): Verifying that each member of the family is a solution of the initial-value problem

$$\begin{gathered} y=cx \\ y\text{'}=c \end{gathered}$$

Then,

$$\begin{gathered} xy\text{'}=y \\ x\left(c\right)=cx \\ cx=cx \end{gathered}$$

Initial value problem:

$$\begin{gathered} y\left(0\right)=0 \\ c\left(0\right)=0 \\ 0=0 \end{gathered}$$

Hence, we find that localid="1668178413158" $\mathit{y}\mathbf{=}\mathit{c}\mathit{x}$ satisfies the differential equation and the initial-value problem by inspection.

3 (b): Determining a region R in the xy-plane for which the differential equation would have a unique solution

Differential equation:

$f=\frac{y}{x}$

Given that,

$\frac{df}{dy}$

Finding

$\frac{\partial f}{\partial y}=\frac{1}{x}$

Therefore, the region of the xy-plane where the differential equation would have a solution is as follows.

For $x>0$or $x<0$, we see that $f$and $\frac{df}{dy}$are not defined at $x=0$.

Hence, the region is any rectangular region not touching the y-axis.

4(c): Determining whether this function is also a solution of the initial-value problem in part (a)

Function at $x=0$and $y=x$

Condition : $y\left(0\right)=0$

At $x=0$, the function that we use is $y=x$. We see that $y\left(0\right)=0$, which satisfies the condition.

So, it is not a solution to the initial value problem in part (a).

We recognize that the piecewise-defined function is not differentiable at $x=0$.

Therefore, the function is not a solution to the initial value problem in part (a).