2.2.16 (a) intial value (b) interval

[karush]

(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?

 

[HOI]

(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, $y= -\frac{1}{\sqrt{2}}$

 

[karush]

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

 

[HOI]

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

 

[karush]

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}

 

[HOI]

The fourth power or a real number is never negative!

 

[karush]

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.

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

 

[HOI]

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 [tex]\frac{dy}{dx}= \frac{x(x^2+ 1)}{4y^3}. You can separate that as $$4y^3dy= x(x^2+ 1)dx$$. 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 [tex]\frac{1}{4}= \frac{1}{4}+ C[tex] so that C= 0. $$y^4= \frac{(x^2+ 1)^2}{4}$$. Taking the fourth root (and making sure we have the right branch) $$y= -\sqrt{\frac{x^2+ 1}}{2}}$$ as your book has.

 

[karush]

what do you mean by branch?

that was a tricky last steps

 

[HOI]

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.