Question

(a) Show that a solution from the family in part (a) of Problem 31 that satisfies y’ = y2, y(1) = 1, is y = 1/(2 − x).

(b) Then show that a solution from the family in part (a) of Problem 31 that satisfies y’ = y2, y(3) = −1, is y = 1/(2 − x).

(c) Are the solutions in parts (a) and (b) the same?

Short answer

a) Proved

b) Proved

c) Solutions in parts (a) and (b) are the same

Step-by-step solution

1(a):  Showing that a solution from the family in part (a) of Problem 31 that satisfies y’ = y2, y(1) = 1, is y = 1/(2 − x)

First let’s show that $y=\frac{1}{2-x}$satisfies $y\text{'}=y^2$

$$\begin{gathered} \left(\frac{1}{2-x}\right)\text{'}={\left(y=\frac{1}{2-x}\right)}^2 \\ \frac{1}{{(2-x)}^2}=\frac{1}{{(2-x)}^2} \end{gathered}$$

It still remains to check if $y\left(1\right)=1:$

$$\begin{gathered} \left(1\right)=\frac{1}{(2-1)} \\ y\left(1\right)=\frac{1}{1} \\ y\left(1\right)=1 \end{gathered}$$

Hence, it is proved that a solution from the family in part (a) of Problem 31 that satisfies y’ = y2, y(1) = 1, is y = 1/(2 − x)

3(b):  Showing that a solution from the family in part (a) of Problem 31 that satisfies y’ = y2, y(3) = −1, is y = 1/(2 − x)

We check if $y\left(3\right)=-1:$

$$\begin{gathered} y\left(3\right)=\frac{1}{(2-3)} \\ y\left(1\right)=-\frac{1}{1} \\ y\left(1\right)=-1 \end{gathered}$$

Remember that a general solution of an first order differential equation is one that involves necessary arbitrary constant, while particular solution is a solution obtained from the general solution by assigning specific value to the arbitrary constant.

Hence, it is proved that a solution from the family in part (a) of Problem 31 that satisfies y’ = y2, y(3) = −1, is y = 1/(2 − x)

Determining the solutions in parts (a) and (b) the same

The conditions for calculating the value of the arbitrary constants can be provided to us in the form of an initial value problem.

We know that general solution is,

$y=-\frac{1}{x+C}$

And if we substitute for $C=-2$, we get

$$\begin{gathered} y=-\frac{1}{x-2} \\ y=\frac{1}{2-x} \end{gathered}$$

So, $y=\frac{1}{2-x}$is one particular solution of given DE.

Therefore, solutions are the same because they belong to the same particular solution.