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2.2.16 (a) intial value (b) interval

karush

Well-known member
Jan 31, 2012
2,886
(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?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
(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

Well-known member
Jan 31, 2012
2,886
$\displaystyle y^4=\frac{x^4}{4} +\frac{ x^2}{2}+ C$
but what about $y^4$
then
$y=\left[\frac{x^4}{4} +\frac{ x^2}{2}\right]^{1/4}$
so where does C go
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
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

Well-known member
Jan 31, 2012
2,886
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:

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
The fourth power or a real number is never negative!
 

karush

Well-known member
Jan 31, 2012
2,886
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.:confused:

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

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
You've lost the negative sign. [tex](0+ 0+ \frac{1}{4})^{1/4}= \frac{1}{\sqrt{2}}[/tex] not [tex]-\frac{1}{\sqrt{2}}[/tex]. 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 [tex]4y^3dy= x(x^2+ 1)dx[/tex]. The left side integrates to [tex]y^4[/tex]. Yes, on the right, you can write this as [tex](x^3+ x)dx[/tex] and integrate as [tex]\frac{x^4}{4}+ \frac{x^2}{2}+ C[/tex]. If we now include the condition that [tex]y(0)= -\frac{1}{\sqrt{2}}[/tex] we have [tex]y^4= \frac{1}{4}= C[/tex] so that [tex]y^4= \frac{x^4}{4}+ frac{x^2}{2}+ \frac{1}{4}[/tex]. 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: [tex]y= -\left(\frac{x^4}{4}+ \frac{x^2}{2}+ \frac{1}{4}\right)^{1/4}[/tex].

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

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

karush

Well-known member
Jan 31, 2012
2,886
what do you mean by branch?

that was a tricky last steps
 
Last edited:

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
As I said, [tex]x^4= a[/tex] 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.