# 2.2.16 (a) intial value (b) interval

• MHB
• karush
In summary, 2.2.16 (a) initial value (b) interval refers to a specific type of mathematical notation used to represent a function or equation. The initial value and interval are important parameters in this notation that provide information about the behavior and characteristics of the function. This notation is commonly used in various fields such as physics, engineering, economics, and computer science. However, it may have limitations in accurately representing complex functions or providing a complete understanding of a function's behavior.
karush
Gold Member
MHB
(a) intial value (b) interval
$\displaystyle y^{\prime}= \frac{x(x^2+1)}{4y^3}, \quad y(0)=-\frac{1}{\sqrt{2}}$
rewrite
$$\displaystyle\d{y}{x}=\frac{x(x^2+1)}{4y^3}$$
$$\displaystyle y^3\, dy=\frac{x(x^2+1)}{4}\, dx$$
integrate
$$\displaystyle\frac{y^4}{4}= \frac{1}{4}\left(\frac{x^4}{4} +\frac{ x^2}{2}\right)$$
$$\displaystyle y^4=\frac{x^4}{4} +\frac{ x^2}{2}$$so far hopefully sorta?

karush said:
(a) intial value (b) interval
$\displaystyle y^{\prime}= \frac{x(x^2+1)}{4y^3}, \quad y(0)=-\frac{1}{\sqrt{2}}$
rewrite
$$\displaystyle\d{y}{x}=\frac{x(x^2+1)}{4y^3}$$
$$\displaystyle y^3\, dy=\frac{x(x^2+1)}{4}\, dx$$
integrate
$$\displaystyle\frac{y^4}{4}= \frac{1}{4}\left(\frac{x^4}{4} +\frac{ x^2}{2}\right)$$
$$\displaystyle y^4=\frac{x^4}{4} +\frac{ x^2}{2}$$so far hopefully sorta?
Almost- you haven't included the "constant of integration" You should have
$$\displaystyle y^4=\frac{x^4}{4} +\frac{ x^2}{2}+ C$$

Now choose C so that when x= 0, $y= -\frac{1}{\sqrt{2}}$

$\displaystyle y^4=\frac{x^4}{4} +\frac{ x^2}{2}+ C$
but what about $y^4$
then
$y=\left[\frac{x^4}{4} +\frac{ x^2}{2}\right]^{1/4}$
so where does C go

That's basic algebra, isn't it? If $y^4= A$ then $y= A^{1/4}$. Since $y^4= \frac{x^4}{4}+ \frac{x^2}{2}$, $y= \left(\frac{x^4}{4}+ \frac{x^2}{2}+ C\right)^{1/4}$.

So C is in inside the radical!

$$y=\left[\frac{x^4}{4} +\frac{ x^2}{2}+ C\right]^{1/4}$$
\begin{align}
y(0)&=\left[\frac{0}{4} +\frac{ 0}{2}+ C\right]^{1/4}=-\frac{1}{\sqrt{2}}\\
&=C^{1/4}=-\frac{1}{\sqrt{2}}\\
&=C=\left(-\frac{1}{\sqrt{2}}\right)^4=-\frac{1}{4}\\
\end{align}

Last edited:
The fourth power or a real number is never negative!

ok the book answer to this is
$$(a)\quad y=-\sqrt{ (x^2 + 1)/2} \quad (c) −∞ < x < ∞$$

doesn't look I am approaching it.

\begin{align} y(0)&=\left[\frac{0}{4} +\frac{ 0}{2}+ C\right]^{1/4}=-\frac{1}{\sqrt{2}}\\ &=C^{1/4}=-\frac{1}{\sqrt{2}}\\ &=C=\left(-\frac{1}{\sqrt{2}}\right)^4=\frac{1}{4}\\ \end{align}

You've lost the negative sign. $$(0+ 0+ \frac{1}{4})^{1/4}= \frac{1}{\sqrt{2}}$$ not $$-\frac{1}{\sqrt{2}}$$. To have a negative after the fourth root you need a negative outside the root.

Let's go back to the beginning. Your differential equation is $$\frac{dy}{dx}= \frac{x(x^2+ 1)}{4y^3}. You can separate that as [tex]4y^3dy= x(x^2+ 1)dx$$. The left side integrates to $$y^4$$. Yes, on the right, you can write this as $$(x^3+ x)dx$$ and integrate as $$\frac{x^4}{4}+ \frac{x^2}{2}+ C$$. If we now include the condition that $$y(0)= -\frac{1}{\sqrt{2}}$$ we have $$y^4= \frac{1}{4}= C$$ so that $$y^4= \frac{x^4}{4}+ frac{x^2}{2}+ \frac{1}{4}$$. To solve for y take the fourth root. But a positive real number has two real fourth roots (and two imaginary). To make sure we have the branch that is negative when x= 0, we have to put the negative sign in explicitely: $$y= -\left(\frac{x^4}{4}+ \frac{x^2}{2}+ \frac{1}{4}\right)^{1/4}$$.

But when integrating $$x(x^2+ 1)dx$$ I would b inclined to use the substition $$u= x^2+ 1$$ so that $$du= 2xdx$$ and $$xdx= \frac{u}{2}du$$. Integrating that gives $$\frac{u^2}{4}+ C= \frac{(x^2+ 1)^2}{4}+ C$$.

So we can also write $$y^4= \frac{(x^2+ 1)^2}{4}+ C$$. For x= 0 we have $$\frac{1}{4}= \frac{1}{4}+ C[tex] so that C= 0. [tex]y^4= \frac{(x^2+ 1)^2}{4}$$. Taking the fourth root (and making sure we have the right branch) $$y= -\sqrt{\frac{x^2+ 1}}{2}}$$ as your book has.

what do you mean by branch?

that was a tricky last steps

Last edited:
As I said, $$x^4= a$$ has, for any positive number a, four roots, two real and two pure imaginary. Those are the four "branches" I was referring to. You want the one that is real and negative.

## 1. What does "2.2.16 (a) intial value (b) interval" refer to?

2.2.16 (a) initial value (b) interval refers to a specific type of notation used in mathematics to represent a function or equation. The "a" and "b" in parentheses refer to specific parameters within the notation.

## 2. How do you determine the initial value and interval in this notation?

The initial value is the value of the function when the input (usually represented as "x") is equal to 0. The interval is the range of values for which the function is defined. To determine these values, you can plug in 0 for x and then find the range of values for which the function is defined.

## 3. What is the significance of the initial value and interval in this notation?

The initial value and interval help to define the behavior and characteristics of the function represented by this notation. They provide important information about the starting point and the range of values for which the function is valid.

## 4. How is this notation used in real-world applications?

This notation is commonly used in physics and engineering to represent physical phenomena such as motion, electricity, and heat. It is also used in economics to model financial data and in computer science to represent algorithms and data structures.

## 5. Are there any limitations to this notation?

Like any mathematical notation, 2.2.16 (a) initial value (b) interval has its limitations. It may not be able to accurately represent complex functions or functions with multiple variables. It also does not provide a complete picture of the behavior of a function and may require additional notation or representations to fully understand its properties.

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